Daily algorithm
those times when you get up early and you work hard; those times when you stay up late and you work hard; those times when don’t feel like working — you’re too tired, you don’t want to push yourself — but you do it anyway. That is actually the dream. That’s the dream. It’s not the destination, it’s the journey. And if you guys can understand that, what you’ll see happen is that you won’t accomplish your dreams, your dreams won’t come true, something greater will. mamba out
Those spare your hard work pay, you do not want to train, when you feel tired but still have to adhere to teeth when that chasing a dream, do not care what the end there, the road to enjoy the process, perhaps you can not dream of success, but there will be more great things will follow. mamba out ~
2020.3.22
Luogu -P3915 decomposition tree
None, except to say too much water and some data directly to the modulo had not be able to sample ac, feel is the right thing but not ac, ah never approach a solution to a problem is a question of the destruction mood legitimate child cut tree cut statistics number
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
const int N = 100005;
int t , n , k;
vector<int> a[N];
int dfs(int now , int last)
{
int ans = 1;
for(int i = 0;i < a[now].size();i ++)
{
if(a[now][i] == last)continue;
int t = dfs(a[now][i],now);
if(t == -1 || t > k)return -1;
else if(t == k)continue;
ans += t;
}
return ans;
}
int main()
{
cin >> t;
while(t--)
{
cin >> n >> k;
for(int i = 0;i <= n;i ++)a[i].clear();
for(int i = 0;i < n - 1;i ++)
{
int x, y ;
scanf("%d %d", &x ,& y);
a[x].push_back(y);
a[y].push_back(x);
}
int ans = dfs(1,1);
if(ans == k){
cout << "YES" << endl;
}else cout << "NO" << endl;
}
return 0;
}
P3956 board
This problem Oh do not say debug two or three hours later, he gave me the whole out, is the test of your mind. And branch judgment is very difficult to judge the need to get control well for each condition, and violence is certainly not pure search was over, the need for pruning, in general is a topic that is too good to be very annoying the
#include <iostream>
#include <cstdio>
#include <vector>
#include <cstring>
#include <queue>
using namespace std;
const int N = 105;
int a[N][N];
int n , m;
int dir[4][2] = {{1,0},{-1,0},{0,1},{0,-1}};
/*
可执行操作
1. 只能走到有颜 色的位置上去
2. 只能向上下左右走
3. 相邻两个格子如果颜色相同不需要花费金币
颜色不同需要花费一个金币
4. 花费两个金币施展魔法让下一个无色格子变成指定颜色
5. 魔法不能连续使用当你离开了施展魔法的格子之后这个格子
会自动变成无色
求从左上角到右下角花费的最小金币？
1 --- 黄色 2 ---- 红色
*/
/*
分支策略：
1. 如果其旁边存在颜色相同的就直接走
2. 如果其旁边存在颜色不同的画一个金币走过去
3. 如果其旁边存在空白格子的话捡起变成根该位置相同颜色
在过去走过之后要恢复这个格子的颜色
*/
struct node{
int x, y ,cost, color;
bool gc;
node(int a,int b,int c,int col): x(a) ,y(b),cost(c),color(col){
this->gc = false;
}
};
int mincost = 0x3f3f3f;
int f[N][N] ; // 记录来到当前位置上的最小金币数
void output()
{
cout << endl;
for(int i = 1;i <= m ;i ++)
{
for(int j = 1;j <= m; j++)
{
printf("%d ",f[i][j]);
}
cout << endl;
}
}
bool inmap(int x,int y)
{
return x >= 1 && x <= m && y >= 1 && y <= m;
}
void bfs(node start)
{
fill(f[0],f[0] + N * N, -1); // 初始化每个点都不能到达
f[1][1] = 0;
queue<node> q;
q.push(start);
while(!q.empty())
{
node now = q.front();
for(int i = 0;i < 4 ;i ++)
{
int nx = now.x + dir[i][0];
int ny = now.y + dir[i][1];
if(now.color == a[nx][ny] && a[nx][ny] > 0 && now.color > 0)
{
if(f[nx][ny] == -1 || f[nx][ny] > f[now.x][now.y])
{
f[nx][ny] = now.cost;
q.push(node(nx,ny,f[nx][ny],now.color));
}
}
if(now.color > 0 && a[nx][ny] > 0 && now.color != a[nx][ny])
{
if(f[nx][ny] == -1 || f[nx][ny] > f[now.x][now.y] + 1)
{
f[nx][ny] = f[now.x][now.y] + 1;
q.push(node(nx,ny,f[nx][ny],a[nx][ny]));
}
}
if(a[nx][ny] == 0)
{
// 当前位置本来既有颜色
if(a[now.x][now.y] > 0)
{
if(f[nx][ny] == -1 || f[nx][ny] > f[now.x][now.y] + 2)
{
f[nx][ny] = f[now.x][now.y] + 2;
q.push(node(nx,ny,f[nx][ny],now.color));
}
}//当前位置是施展魔法得来的
if(a[now.x][now.y] == 0)
{
continue;
}
}
}
q.pop();
}
//output();
}
void input()
{
cin >> m >> n;
for(int i = 0;i < n ;i ++)
{
int x, y , c;
scanf("%d %d %d",&x , &y, &c);
a[x][y] = c == 1 ? 1 : 2;
}
}
void work()
{
bfs(node(1,1,0,a[1][1]));
}
int main()
{
input();
/*for(int i = 1;i <= m ;i ++)
{
for(int j = 1;j <= m ;j ++)
{
printf("%d ",a[i][j]);
}
printf("\n");
} */
work();
cout << f[m][m];
return 0;
}