1005: Biggest Number
时间限制: 1 内存限制: 128 MB
题目描述
You have a maze with obstacles and non-zero digits in it:
You can start from any square, walk in the maze, and finally stop at some square. Each step, you may only walk into one of the four neighbouring squares (up, down, left, right) and you cannot walk into obstacles or walk into a square more than once. When you finish, you can get a number by writing down the digits you encounter in the same order as you meet them. For example, you can get numbers 9784, 4832145 etc. The biggest number you can get is 791452384, shown in the picture above.
Your task is to find the biggest number you can get.
输入
There will be at most 25 test cases. Each test begins with two integers R and C (2<=R,C<=15, R*C<=30), the number of rows and columns of the maze. The next R rows represent the maze. Each line contains exactly C characters (without leading or trailing spaces), each of them will be either '#' or one of the nine non-zero digits. There will be at least one non-obstacle squares (i.e. squares with a non-zero digit in it) in the maze. The input is terminated by a test case with R=C=0, you should not process it.
输出
For each test case, print the biggest number you can find, on a single line.
样例输入
3 7
##9784#
##123##
##45###
0 0样例输出
791452384参考了CSDN大佬的强剪枝（膜拜.jpg）
BFS来获取任意点的可达最长路径，用于剪枝，DFS实现最大值的获取与更新，本题用string实现字符串拼接与大小比较，当然也可用strcmp与strcat函数实现，较麻烦一点。
AC Code:
#include <iostream>
#include<cstdint>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<climits>
#include<map>
#include<queue>
using namespace std;
static const int dir[4][2] = {{1, 0}, {-1, 0}, {0, -1}, {0, 1}};
bool vis[20][20], used[20][20];
int n, m;
char maps[20][20], str[35];
string res, s1, s2;
struct Point{
int x, y;
};
bool is_bound(int x, int y){
return (x >= 0 && x < n && y >= 0 && y < m && maps[x][y] != '#');
}
int path(int x, int y){ //获取x,y点可达最长路径
Point p1, p2;
int tail = 0;
p1.x = x, p1.y = y;
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
used[i][j] = vis[i][j];
}
}
queue<Point>q;
q.push(p1);
while(!q.empty()){
p1 = q.front();
q.pop();
for(int i = 0; i < 4; i++){
p2.x = p1.x + dir[i][0];
p2.y = p1.y + dir[i][1];
if(is_bound(p2.x, p2.y) && !used[p2.x][p2.y]){
used[p2.x][p2.y] = true;
str[tail++] = maps[p2.x][p2.y];
q.push(p2);
}
}
}
str[tail] = '\0';
return tail;
}
void dfs(int x, int y, string s){
int len = path(x, y);
if(s.length() > res.length() || (s.length() == res.length() && s > res)){
res = s; //最大值更新
}
if(len + s.length() < res.length()) return; //长度比res小的剪枝
if(len + s.length() == res.length()){ //长度相等，大小可能超过res
sort(str, str + len); //获取路径最大值用于剪枝
s2 = s;
for(int i = len - 1; i >= 0; i--){ //s2最大值
s2 += str[i];
}
if(s2 < res) return; //长度相等，字符串小的剪枝
}
for(int i = 0; i < 4; i++){
int fx = x + dir[i][0];
int fy = y + dir[i][1];
if(is_bound(fx, fy) && !vis[fx][fy]){
vis[fx][fy] = true;
dfs(fx, fy, s + maps[fx][fy]);
vis[fx][fy] = false; //回溯
}
}
}
int main(){
while(scanf("%d%d", &n, &m) != EOF){
if(n == 0 && m == 0) break;
memset(vis, false, sizeof(vis));
res = "";
for(int i = 0; i < n; i++){
scanf("%s", maps[i]);
}
for(int i = 0; i < n; i++){
for(int j = 0; j < m; j++){
if(maps[i][j] != '#'){
vis[i][j] = true;
s1 = "";
dfs(i, j, s1 + maps[i][j]);
vis[i][j] = false;
}
}
}
// cout<<res<<endl;
printf("%s\n", res.c_str());
}
return 0;
}