输入样例 5 0 2 3 0 8 2 0 0 0 5 3 0 0 1 0 0 0 1 0 1 8 5 0 1 0 1 5

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5+10;
int g[100][100],a,b,n,h[100];
//5
//0 2 3 0 8
//2 0 0 0 5
//3 0 0 1 0
//0 0 1 0 1
//8 5 0 1 0
//1 5

int dfs(int x){
	h[x] = 1;
	for(int i=1;i<=n;i++){
		if(h[i]==0&&g[x][i]>0){
			h[i]=1;
			if(g[a][x]+g[x][i]<g[a][i]){
				g[a][i]=g[a][x]+g[x][i];
			}
			dfs(i);
			h[i]=0;
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			printf("%d ",g[i][j]);
		}
		printf("\n");
	}
	printf("\n");
	return 0;
}


int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			scanf("%d",&g[i][j]);
			if(g[i][j]==0){
				g[i][j]=1e9;
			}
		}
	}
	scanf("%d%d",&a,&b);
	dfs(a);
	printf("%d",g[a][b]);
	return 0;
}

向量容器

# P5318 【深基18.例3】查找文献

### 1213：八皇后问题