1: chessboard
In a checkerboard given shape (the shape may be irregular) placed above the pieces, pieces no difference. The requirements of any two pieces can not be placed in the same row or the same column in the board when the display is required, program for solving a given board size and shape, placing the k pieces of all possible placement scheme C.
This problem should pay attention to any two pieces on the chessboard are not the same column or the same row, by this sentence we can start from the problem into the first rows, each only a place, and not in the same column with the above pieces then we just follow the traversal of each row or column or a row, then DFS find all possible.
If traversed by row, we just need to mark what columns are let off pieces.
This question is simple lines but according to the method of search is very clever
Code:
#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
#define fo(i,a,b) for(int i = a; i < b; i++) //宏定义 加快函数速度并简化函数写法
#define met(a,b) memset(a,0,sizeof(a));
char map[9][9];
int vis[9]; // 因为每次都是按行搜索，所以一定不在同一行，所以只需要标记列。
int n, m, sum, num;
void print(){ // 棋盘输入
met(map, 0);
met(vis, 0);
sum = 0, num = 0;
fo(i, 0, n)
fo(j, 0, n)
cin>>map[i][j];
}
void dfs(int x,int num){
if(num==m)
{
sum++;
return ; //当棋子全部放入棋盘后，返回最后一层的上一层；
}
fo(i, x, n)
fo(j, 0, n){
if(map[i][j]=='#'&&!vis[j]){
vis[j] = 1;
dfs(i+1,num+1); // 递归节点
vis[j] = 0; // 回溯节点
}
}
}
int main()
{
while(cin>>n>>m&&n!=-1&&m!=-1){
print();
dfs(0,0);
cout<<sum<<endl;
}
return 0;
} 2. Dungeon Master
BFS just from ordinary two-dimensional into a three-dimensional, that is, more than two directions, up and down,
#include <stdio.h>
#include <iostream>
#include <string>
#include <string.h>
#include <math.h>
#include <queue>
using namespace std;
#define met(a,b) memset(a,b,sizeof(a))
#define for(i,a,b) for(int i = a; i < b; i++)
struct node{
int x,y,z;
node(int x = 0, int y = 0, int z = 0):x(x), y(y), z(z){}
};
queue <node> q;
int vis[33][33][33];
int step[33][33][33];
char map[33][33][33];
int nex[6][3] = {{0, 1, 0}, {1, 0, 0}, {0, 0, 1}, {0, 0, -1}, {0, -1, 0}, {-1, 0, 0}};
int L, R, C, sx, sy, sz, count1;
void init(){ // 输入
met(vis, 0);
met(map, 0);
met(step, 0);
count1 = 0;
while(!q.empty()) q.pop();
for(l, 0, L)
for(i, 0, R)
for(j, 0, C){
cin>>map[i][j][l];
if(map[i][j][l] == 'S') {sx = i; sy = j, sz = l;}
}
}
void bfs(){
node a;
a.x = sx; a.y = sy; a.z = sz;
q.push(a);
while(!q.empty()){
node b = q.front();
q.pop();
for(i, 0, 6){
a.x = b.x + nex[i][0];
a.y = b.y + nex[i][1];
a.z = b.z + nex[i][2];
if((map[a.x][a.y][a.z] == '.' || map[a.x][a.y][a.z] == 'E') && !vis[a.x][a.y][a.z]){
vis[a.x][a.y][a.z] = 1;
step[a.x][a.y][a.z] = step[b.x][b.y][b.z] + 1;
if(map[a.x][a.y][a.z] == 'E'){count1 = 1; printf("Escaped in %d minute(s).\n",step[a.x][a.y][a.z]); return; }
q.push(a);
}
}
}
return ;
}
int main(){
while(cin>>L>>R>>C,L, R, C){
init();
bfs();
if(!count1) {cout<<"Trapped!"<<endl;}
}
return 0;
}3.Catch That Cow
Simple bfs
#include <stdio.h>
#include <iostream>
#include <string>
#include <string.h>
#include <queue>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
struct node{
int x,step;
node(int x=0, int step=0):x(x), step(step){}
};
int N,K;
int vis[1000002];
void bfs(){
memset(vis, 0, sizeof(vis));
queue <node> q;
node a;
a.x = N;
a.step=0;
q.push(a);
vis[a.x] = 1;
while(!q.empty()){
node r = q.front();
q.pop();
for(int i = 0; i < 3; i++){
if(i==0){
a.x = r.x + 1;
if(!vis[a.x]&&a.x>=0&&a.x<=100000){
a.step=r.step+1,q.push(a),vis[a.x]=1;
if(a.x==K) {cout<<a.step<<endl;return;}
}
}
if(i==1){
a.x = r.x - 1;
if(!vis[a.x]&&a.x>=0&&a.x<=100000) {
a.step=r.step+1,q.push(a),vis[a.x]=1;
if(a.x==K) {cout<<a.step<<endl;return;}
}
}
if(i==2){
a.x = r.x * 2;
if(!vis[a.x]&&a.x>=0&&a.x<=100000){
a.step=r.step+1,q.push(a),vis[a.x]=1;
if(a.x==K) {cout<<a.step<<endl;return;}
}
}
}
}
return ;
}
int main(){
cin >> N >> K ;
if(N==K) cout<<"0"<<endl;
else bfs();
return 0;
} 4. Fliptile (+ problem binary search switch) may change each position of the surrounding four directions, if the rows or columns of view, only the line above the line to the next line can affect, can simplify the problem, we can enumerate all cases the first line and the second line from the start, so the first line of white. Until then the last line, if the last line is white, the situation will be reversed array save up update reversals smaller, the final output.
#include <iostream>
#include <string>
#include <cstring>
#include <queue>
using namespace std;
#define met(a,b) memset(a,b,sizeof(a))
#define fo(i,a,b) for(int i = a; i < b; i++)
int map[20][20];
int cal[20][20];
int Min[20][20];
int n,m;
int nxt[5][2] = {{0,0},{0,1},{0,-1},{1,0},{-1,0}};
int getc(int x,int y){ //(x,y)的状态由本身的黑白 + 周围五个的翻转状态决定
int t = map[x][y];
for(int i = 0;i < 5;i ++){
int nx = x + nxt[i][0];
int ny = y + nxt[i][1];
if(nx < 1 || nx > n || ny < 1 || ny > m) continue;
t += cal[nx][ny];
}
return t%2;
}
int dfs(){
for(int i = 2; i <= n; i++)
for(int j = 1;j <= m;j ++)
if(getc(i-1,j)) cal[i][j] = 1; //如果上方为黑色，必须要翻转
for(int i = 1; i <= m; i++) //最后一行全白
if(getc(n,i)) return -1;
int t = 0;
for(int i = 1;i <= n;i ++)
for(int j = 1;j <= m;j ++)
t += cal[i][j];
return t;
}
int main()
{
while(cin>>n>>m){
for(int i = 1;i <= n; i++)
for(int j = 1;j <= m; j++)
cin>>map[i][j];
int flag = 0, ans = -1;
for(int i = 0;i < 1<<m; i++){ //第一行 1<<m种状态，二进制从0开始，字典序从小到大
memset(cal,0,sizeof(cal));
for(int j = 1;j <= m; j++) //利用二进制枚举第一行所有的情况
cal[1][m-j+1] = i>>(j-1) & 1;
int cont = dfs();
if(cont >= 0 &&(cont < ans || ans<0)){ //翻转次数最少
flag = 1;
ans = cont;
memcpy(Min,cal,sizeof(cal));
}
}
if(!flag) cout<<"IMPOSSIBLE"<<endl;
else{
for(int i = 1;i <= n;i ++){
for(int j = 1;j <= m;j ++){
if(j != 1) cout<<" ";
cout<<Min[i][j];
}
cout<<endl;
}
}
}
return 0;
}5.Find The Multiple
long long answer may be able to save the largest, simple BFS
#include <stdio.h>
#include <string>
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
#define ll long long
int x;
void bfs(){
queue<ll> q;
q.push(1);
while(!q.empty()){
ll b = q.front();
q.pop();
if(b % x == 0) {cout<<b<<endl; return;}
for(int i = 0; i < 2; i++){
if(i == 0){ ll a = b*10 + 0; q.push(a);}
else { ll a = b*10 + 1; q.push(a);}
}
}
}
int main(){
while(cin>>x,x){
bfs();
}
return 0;
}6.Prime Path
This is still the prime number after a number of changes to a prime number, each time a change, and change. Seeking a minimum number of operations that BFS certainly get away.
Can advance all four-digit prime numbers play table, then you can directly inquired.
Each time through all the circumstances of a bit, the number of all eligible to join the queue.
#include <iostream>
#include <string.h>
#include <stdio.h>
#include <queue>
#include <math.h>
#include <algorithm>
using namespace std;
#define met(a,b) memset(a,b,sizeof(a))
#define fo(i,a,b) for(int i = a; i < b; i++)
int vis[10000];
int A[10000]={0};
int x, y;
bool flag;
typedef pair <int, int> P;
bool sushu(int a){
for(int i = 2; i < a/2+1 ;i++)
if(a%i==0) return false;
return true;
}
void FindA(){
for(int i = 999; i<9999;i+=2){
if(sushu(i)){
A[i] = 1;
}
}
return ;
}
void solve(int sum,int &A, int &B, int &C, int &D){
D = sum / 1000;
C = sum /100 %10;
B = sum /10 %10;
A = sum %10;
return ;
}
void bfs(){
flag=false;
for(int i = 1000; i <= 9999; i+=1)
vis[i] = 0;
queue <P> q;
q.push(P(x,0));
vis[x] = 1;
while(!q.empty()){
P p = q.front();
q.pop();
int ONE,TEN,HUN,TND;
solve(p.first,ONE,TEN, HUN, TND);
if(p.first == y) {cout<<p.second<<endl; flag=true; return;}
//qian
for(int j = 1; j <= 9; j++){
P pp;
pp.first = j*1000 + HUN*100 + TEN*10 + ONE;
// if(pp.first==y) {cout << p.second+1 <<endl;flag=true;return;}
if(A[pp.first]&&!vis[pp.first]) {pp.second=p.second+1;q.push(pp);vis[pp.first]=1;}
}
//bai
for(int j = 0; j <= 9; j++){
P pp;
pp.first = TND*1000 + j*100 + TEN*10 + ONE;
// if(pp.first==y) {cout << p.second+1 <<endl;flag=true;return;}
if(A[pp.first]&&!vis[pp.first]) {pp.second=p.second+1;q.push(pp);vis[pp.first]=1;}
}
//shi
for(int j = 0; j <= 9; j++){
P pp;
pp.first = TND*1000 + HUN*100 + j*10 + ONE;
// if(pp.first==y) {cout << p.second+1 <<endl;flag=true;return;}
if(A[pp.first]&&!vis[pp.first]) {pp.second=p.second+1;q.push(pp);vis[pp.first]=1;}
}
//ge
for(int j = 0; j <= 9; j++){
P pp;
pp.first = TND*1000 + HUN*100 + TEN*10 + j;
// if(pp.first==y) {cout << p.second+1 <<endl;flag=true;return;}
if(A[pp.first]&&!vis[pp.first]) {pp.second=p.second+1;q.push(pp);vis[pp.first]=1;}
}
}
return ;
}
int main(){
int T;
FindA();
cin>>T;
while(T--)
{
cin>>x>>y;
bfs();
if(flag==false)
cout << "Impossible\n";
}
return 0;
}7. Shuffle'm Up
Bottommost card is S [0];
Violence analog, then there have been added character map array, if there are duplicate, then exits the output 1;
#include <cstdio>
#include <iostream>
#include <queue>
#include <string>
#include <map>
#include <set>
using namespace std;
int main(){
int T;
cin>>T;
for(int i = 0; i < T; i++){
map <string,int> S;
int m,cnt = 0;
cin>>m;
string s,s1,s2,s3=""; // s + s1 -> s2
cin>>s>>s1>>s2;
while(1){
s3="";
cnt++;
for(int i = 0; i < m; i++){
s3 =s3 + s1[i] + s[i];
}
if(S[s3]) {cnt = -1;break;}
if(s3==s2) {break;}
else{
S[s3] = 1;
s = s3.substr(0,m);
s1 = s3.substr(m,m);
}
}
cout<<i+1<<" "<<cnt<<endl;
}
return 0;
}8.Pots
Pour problem, the key is that all the water bottles seen as a state, then pour on the line simulation.
This problem also stored at the last pour values, reverse search path, reverse output path after save.
#include <stdio.h>
#include <string>
#include <iostream>
#include <queue>
#include <cstring>
using namespace std;
#define met(a,b) memset(a,b,sizeof(a))
#define fo(i,a,b) for(int i = a; i < b; i++)
struct node{
int a,b,step;
};
int vis[110][110];
node l[110][110];
int A,B,C,flag = 0;
node cnt;
void bfs(){
queue <node> q;
node nx;
nx.a = 0;
nx.b = 0;
nx.step = 0;
l[nx.a][nx.b].a = 0;
l[nx.a][nx.b].b = 0;
q.push(nx);
while(!q.empty()){
node r = q.front();
q.pop();
if(r.a==C||r.b==C) {flag = 1; cout<<r.step<<endl; cnt.a = r.a; cnt.b = r.b; return ;}
if(r.a!=A){ // 装满 A
nx.a = A;
nx.b = r.b;
if(!vis[nx.a][nx.b]){
vis[nx.a][nx.b] = 1;
l[nx.a][nx.b].a = r.a;
l[nx.a][nx.b].b = r.b;
nx.step = r.step + 1;
q.push(nx);
}
}
if(r.b!=B){ // 装满 B
nx.a = r.a;
nx.b = B;
if(!vis[nx.a][nx.b]){
vis[nx.a][nx.b] = 1;
l[nx.a][nx.b].a = r.a;
l[nx.a][nx.b].b = r.b;
nx.step = r.step + 1;
q.push(nx);
}
}
if(r.a){ // 倒掉 A
nx.a = 0;
nx.b = r.b;
if(!vis[nx.a][nx.b]){
vis[nx.a][nx.b] = 1;
l[nx.a][nx.b].a = r.a;
l[nx.a][nx.b].b = r.b;
nx.step = r.step + 1;
q.push(nx);
}
}
if(r.b){ // 倒掉 B
nx.b = 0;
nx.a = r.a;
if(!vis[nx.a][nx.b]){
vis[nx.a][nx.b] = 1;
l[nx.a][nx.b].a = r.a;
l[nx.a][nx.b].b = r.b;
nx.step = r.step + 1;
q.push(nx);
}
}
if(r.a&&r.b!=B){ // 将 A -> B
int x = r.a + r.b;
if(x<=B) {nx.b = x; nx.a = 0;}
else {nx.b = B; nx.a = x - B;}
if(!vis[nx.a][nx.b]){
vis[nx.a][nx.b] = 1;
l[nx.a][nx.b].a = r.a;
l[nx.a][nx.b].b = r.b;
nx.step = r.step + 1;
q.push(nx);
}
}
if(r.b&&r.a!=A){ // 将 B - > A
int x = r.a + r.b;
if(x<=A) {nx.a = x; nx.b = 0;}
else {nx.a = A; nx.b = x - A;}
if(!vis[nx.a][nx.b]){
vis[nx.a][nx.b] = 1;
l[nx.a][nx.b].a = r.a;
l[nx.a][nx.b].b = r.b;
nx.step = r.step + 1;
q.push(nx);
}
}
}
cout<<"impossible\n";
return ;
}
void print(){
if(flag){
node x[100000];
int i = cnt.a;
int j = cnt.b;
int k = 1;
x[0].a = i;
x[0].b = j;
while(i!=0||j!=0){
x[k].a = l[i][j].a;
x[k].b = l[i][j].b;
int t = i;
i = l[i][j].a;
j = l[t][j].b;
k++;
}
for(int i = k-1; i>0; i--){
// cout<<x[i].a<<" "<<x[i].b<<" "<<x[i-1].a<<" "<<x[i-1].b<<endl;
if(x[i-1].a-x[i].a==0&&x[i-1].b-x[i].b!=0){
if(x[i-1].b==B){
cout<<"FILL(2)\n";
}
else if(x[i-1].b==0)
cout<<"DROP(2)\n";
}
else if(x[i-1].a-x[i].a!=0&&x[i-1].b-x[i].b==0){
if(x[i-1].a==A){
cout<<"FILL(1)\n";
}
else if(x[i-1].a==0)
cout<<"DROP(1)\n";
}
else if(x[i-1].a-x[i].a!=0&&x[i-1].b-x[i].b!=0){
if(x[i-1].a-x[i].a>0&&x[i].b-x[i-1].b>0) // b -> a;
cout<<"POUR(2,1)\n";
else if(x[i].a-x[i-1].a>0&&x[i-1].b-x[i].b>0) // a - > b;
cout<<"POUR(1,2)\n";
}
}
}
}
int main(){
while(cin>>A>>B>>C){
flag = 0;
memset(vis,0,sizeof(vis));
memset(l,0,sizeof(l));
vis[0][0] = 1;
bfs();
print();
}
return 0;
}
9.Fire Game
Determines whether the burn straw, the minimum output time, there are two start point, the number of input save up grass, and BFS, the number of grass was determined to BFS traversal, and a total amount of grass, and if is not equal, the output 1;
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
using namespace std;
#define met(a,b) memset(a,b,sizeof(a))
#define fo(i,a,b) for(int i = a; i < b; i++)
char map[15][15];
int vis[15][15];
int nxt[4][2] = {0,1,1,0,0,-1,-1,0};
int n,m;
struct node{
int x,y,step;
node(int x = 0,int y = 0, int step = 0):x(x),y(y),step(step){}
}Arr[100];
int k;
int minnt = 0;
void init(){
k = 0;
met(Arr,0);
fo(i,0,n){
fo(j,0,m)
{
cin>>map[i][j];
if(map[i][j] == '#'){ Arr[k].x = i; Arr[k].y = j; k++;}
}
}
}
int bfs(int x, int y, int x1, int y1){
met(vis,0);
vis[x][y] = 1;
vis[x1][y1] = 1;
int cnt = 0;
queue <node> q;
node nx, nx1;
nx.x = x; nx.y = y;
nx1.x = x1; nx1.y = y1;
q.push(nx);
q.push(nx1);
if(x==x1&&y==y1) cnt = 1;
else cnt = 2;
while(!q.empty()){
node r = q.front();
q.pop();
for(int i=0;i<4;i++){
nx.x = r.x + nxt[i][0];
nx.y = r.y + nxt[i][1];
if(nx.x<0||nx.x>=n||nx.y<0||nx.y>=m) continue;
if(!vis[nx.x][nx.y]&&map[nx.x][nx.y]=='#'){
vis[nx.x][nx.y] = 1;
nx.step = r.step + 1;
minnt = nx.step;
cnt++;
q.push(nx);
}
}
}
if(cnt == k){
return 1;
}
else return 0;
}
int main(){
int T,count = 0;
cin>>T;
while(T--){
count++;
cin>>n>>m;
init();
// fo(i,0,k) cout<<Arr[i].x<<" "<<Arr[i].y<<endl;
int minn = -1;
for(int i = 0; i < k; i++)
for(int j = i; j < k; j++){
if(bfs(Arr[i].x,Arr[i].y,Arr[j].x,Arr[j].y)&&(minn>minnt||minn==-1))
minn = minnt;
}
if(k<=2) minn = 0;
cout<<"Case "<<count<<": "<<minn<<endl;
}
return 0;
} 10.Fire!
First of fire were BFS, then record each of the firing spread to time, and then people were BFS, to judge whether a certain point in time earlier than when the fire BFS, if it can not be better than the fire early.
#include <iostream>
#include <string.h>
#include <queue>
using namespace std;
#define fo(i,a,b) for(int i = a; i < b; i++)
#define met(a,b) memset(a,b,sizeof(a));
typedef struct Point{
int x,y;
Point(int x=0, int y=0):x(x),y(y){}
}Point;
int nex[4][2]={0, 1, 1, 0, 0, -1, -1, 0};
char map[1010][1010];
int vis[1010][1010];
int timef[1010][1010],timej[1010][1010];
int n, m, x, y, x1, y1;
queue <Point> q;
void Fbfs(){
met(vis, 0);
met(timef, 0);
while(!q.empty()){
Point a = q.front();
vis[a.x][a.y] = 1;
q.pop();
fo(i, 0, 4){
Point b;
b.x = a.x + nex[i][0];
b.y = a.y + nex[i][1];
if(map[b.x][b.y] =='.'&&!vis[b.x][b.y]&&b.x>=0&&b.x<n&&b.y>=0&&b.y<m){
vis[b.x][b.y] = 1;
timef[b.x][b.y] = timef[a.x][a.y] + 1; // 向外扩展的层数
q.push(b);
}
}
}
return;
}
int Jbfs(){
met(vis, 0);
met(timej, 0);
vis[x][y] = 1;
Point a;
a.x = x;
a.y = y;
q.push(a);
while(!q.empty()){
Point b = q.front();
q.pop();
fo(i, 0, 4){
a.x = b.x + nex[i][0];
a.y = b.y + nex[i][1];
if(a.x<0||a.y<0||a.x>=n||a.y>=m){
printf("%d\n",timej[b.x][b.y]+1);
return 0;
}
else if(map[a.x][a.y]=='.'&&(timef[a.x][a.y]>timej[b.x][b.y]+1||timef[a.x][a.y]==0)&&!vis[a.x][a.y]){
vis[a.x][a.y] = 1; //注意判断条件 (timef[a.x][a.y]>timej[b.x][b.y]+1||timef[a.x][a.y]==0) 火走不到或者人走到这一步时，火走不到
timej[a.x][a.y] = timej[b.x][b.y] + 1; // 向外扩展的层数
q.push(a);
}
}
}
printf("IMPOSSIBLE\n");
return 0;
}
void print()
{
while(!q.empty())
q.pop();
met(map, 0);
fo(i, 0, n){
fo(j, 0, m)
{
scanf("%c",&map[i][j]);
if(map[i][j]=='F'){
Point c;
c.x = i;
c.y = j;
q.push(c); //将所有火的坐标都进队，同时进行BFS向外层层扩展
}
if(map[i][j]=='J'){
x = i;
y = j;
}
}
getchar();
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--){
scanf("%d %d", &n, &m);
getchar();
print();
Fbfs();
Jbfs();
}
return 0;
}11. maze
Simple BFS, but requires an output path, is the search output path a lot of trouble, but this question can search backwards, would facilitate that.
#include <iostream>
#include <string>
#include <string.h>
#include <queue>
using namespace std;
#define met(a,b) memset(a,0,sizeof(a))
#define fo(i,a,b) for(int i = a; i < b; i++)
int nex[4][2] = {-1,0,0,-1,0,1,1,0};
int map[5][5];
int vis[5][5];
struct node{
int x,y;
}n[5][5];
void init(){
met(vis, 0);
met(map, 0);
fo(i, 0, 5)
fo(j, 0, 5)
cin>>map[i][j];
}
void bfs(){
queue <node> q;
node a;
a.x = 4; a.y = 4;
q.push(a);
while(!q.empty()){
node b = q.front(); q.pop();
fo(i, 0, 4){
a.x = b.x + nex[i][0];
a.y = b.y + nex[i][1];
if(map[a.x][a.y]==0&&!vis[a.x][a.y]&&a.x>=0&&a.x<5&&a.y>=0&&a.y<5){
q.push(a);
vis[a.x][a.y] = 1;
n[a.x][a.y].x = b.x;
n[a.x][a.y].y = b.y;
}
if(a.x==0&&a.y==0) return;
}
}
}
int main()
{
init();
bfs();
int i = 0, j = 0;
while(1){
cout<<"("<<i<<", "<<j<<")"<<endl;
if(i==4&&j==4) return 0;
int ii = i;
i = n[i][j].x;
j = n[ii][j].y;
}
return 0;
}Seeking the number 12.DFS communication blocks.
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
using namespace std;
char map[110][110];
int vis[110][110];
int nxt[8][2] = {{0,1},{1,1},{1,0},{1,-1},{0,-1},{-1,-1},{-1,0},{-1,1}};
int n,m;
void init(){
memset(vis,0,sizeof(vis));
memset(map,0,sizeof(map));
for(int i = 0; i < n; i++){
scanf("%s",map[i]);
getchar();
}
}
void dfs(int x, int y){
for(int i = 0; i < 8; i++){
int nx = x + nxt[i][0];
int ny = y + nxt[i][1];
if(!vis[nx][ny]&&nx>=0&&nx<n&&ny>=0&&ny<m){
vis[nx][ny] = 1;
if(map[nx][ny]=='@') dfs(nx, ny);
}
}
}
int main(){
while(cin>>n>>m,n,m){
getchar();
init();
int cnt = 0;
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(!vis[i][j]&&map[i][j]=='@'){
vis[i][j] = 1;
dfs(i,j);
cnt++;
}
}
}
cout<<cnt<<endl;
}
return 0;
}13. Cola
Pouring problem. The volume of water is used as a three bottle state, then pour BFS simulation. (Considering only two bottles when pouring)
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <queue>
using namespace std;
struct node{
int s,n,m,step;
};
int S,N,M;
int vis[105][105][105];
void bfs(){
memset(vis,0,sizeof(vis));
vis[S][0][0] = 1;
node s;
s.s = S; s.n = 0; s.m = 0; s.step = 0;
queue <node> q;
q.push(s);
while(!q.empty()){
node r = q.front();
q.pop();
if(r.s==S/2&&r.m==S/2){ cout<<r.step<<endl; return;}
if(r.s&&r.m!=M){ // s -> M;
int c = r.m + r.s;
if(c<=M) {s.s = 0; s.m = r.m + r.s; s.n = r.n;}
else {s.s = c - M; s.m = M; s.n = r.n;}
if(!vis[s.s][s.n][s.m])
{
s.step = r.step + 1;
vis[s.s][s.n][s.m] = 1;
q.push(s);
}
}
if(r.s&&r.n!=N){ // s -> N;
int c = r.n + r.s;
if(c<=N) {s.s = 0; s.n = r.n + r.s; s.m = r.m;}
else {s.s = c - N; s.n = N; s.m = r.m;}
if(!vis[s.s][s.n][s.m])
{
s.step = r.step + 1;
vis[s.s][s.n][s.m] = 1;
q.push(s);
}
}
if(r.n&&r.s!=S){ // N -> S ;
s.s = r.n+r.s; s.n = 0; s.m = r.m;
if(!vis[s.s][s.n][s.m])
{
s.step = r.step + 1;
vis[s.s][s.n][s.m] = 1;
q.push(s);
}
}
if(r.n&&r.m!=M){ // N - > M;
int c = r.m + r.n;
if(c<=M) {s.s = r.s; s.m = r.m + r.n; s.n = 0;}
else {s.s = r.s; s.m = M; s.n = c - M;}
if(!vis[s.s][s.n][s.m])
{
s.step = r.step + 1;
vis[s.s][s.n][s.m] = 1;
q.push(s);
}
}
if(r.m&&r.s!=S){ // M - > S
s.s = r.s + r.m; s.m = 0; s.n = r.n;
if(!vis[s.s][s.n][s.m])
{
s.step = r.step + 1;
vis[s.s][s.n][s.m] = 1;
q.push(s);
}
}
if(r.m&&r.n!=N){ //M - > N;
int c = r.m + r.n;
if(c<=N) {s.s = r.s; s.n = r.m + r.n; s.m = 0;}
else {s.s = r.s; s.n = N; s.m = c - N;}
if(!vis[s.s][s.n][s.m])
{
s.step = r.step + 1;
vis[s.s][s.n][s.m] = 1;
q.push(s);
}
}
}
cout<<"NO"<<endl;
return;
}
int main(){
while(scanf("%d%d%d",&S,&N,&M),S||N||M){
if(N>M){
int t = N;
N = M;
M = t;
}
if(S%2!=0){
cout<<"NO\n";
continue;
}
else
bfs();
}
return 0;
} 14. twice the BFS, two people were recorded KFC arrival time, and then obtains a two KFC reach the same minimum time.
#include <stdio.h>
#include <iostream>
#include <queue>
#include <string>
#include <string.h>
using namespace std;
#define met(a,b) memset(a,b,sizeof(a))
#define for(i,a,b) for(int i=a;i<b;i++)
#define INF 1e9
struct Point{
int x,y;
Point(int x=0, int y=0):x(x),y(y){}
};
char map[202][202];
int vis[202][202];
int nex[4][2] = {0, 1, 1, 0, 0, -1, -1, 0};
int x,y,xx,yy,n,m;
int a[202][202];
int b[202][202];
queue <Point> q;
void print(){
met(map, 0);
met(vis, 0);
met(a, 0);
met(b, 0);
for(i, 0, n){
for(j, 0, m){
scanf("%c",&map[i][j]);
if(map[i][j] == 'Y')
x = i, y = j;
else if(map[i][j]=='M')
xx = i, yy = j;
}
getchar();
}
// for(i, 0, n)
// {
// for(j, 0, m)
// cout<<map[i][j];
// cout<<endl;
// }
}
void Bfs(int x,int y,int num[202][202]){
vis[x][y] = 1;
Point t;
t.x = x;
t.y = y;
q.push(t);
while(!q.empty()){
Point r = q.front();
q.pop();
for(i, 0, 4){
t.x = r.x + nex[i][0];
t.y = r.y + nex[i][1];
if(map[t.x][t.y] == '#'||t.x<0||t.x>=n||t.y<0||t.y>=m||vis[t.x][t.y])
continue;
else{
vis[t.x][t.y] = 1;
q.push(t);
num[t.x][t.y] = num[r.x][r.y] + 1;
}
}
}
}
void slove(){
int minn = INF;
for(i, 0, n)
{
for(j, 0, m){
if(map[i][j] == '@'&&a[i][j]!=0&&b[i][j]!=0)
minn = min(minn, a[i][j]+b[i][j]);
}
}
cout<<minn*11<<endl;
}
int main()
{
while(scanf("%d%d",&n,&m)!=EOF){
getchar();
print();
Bfs(x, y, a);
met(vis, 0);
Bfs(xx, yy, b);
slove();
}
return 0;
}
