题目连接
题意:最短路径Floyd裸题
注意:数组初始化mp[i][i] = 0;还有两个点之间可能有多条路
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<time.h>
#include<set>
#include<stack>
#include<vector>
#include<map>
#define pi acos(-1)
#define maxn 111111
#define maxm 11111
#define INF 0x3F3F3F3F
#define eps 1e-8
#define pb push_back
#define mem(a) memset(a,0,sizeof a)
using namespace std;
const long long mod = 1000000007;
int n, m;
int s, t;
int mp[1111][1111];
void init(void) {
memset(mp, INF, sizeof mp);
for (int i = 0; i <= n; i++) {
mp[i][i] = 0;
}
}
void Floyd() {
for (int k = 0; k < n; k++) {
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]);
}
}
}
}
int main() {
while (scanf("%d%d", &n, &m) != EOF) {
init();
int a, b, x;
for (int i = 1; i <= m; i++) {
scanf("%d%d%d", &a, &b, &x);
if(x < mp[a][b])
mp[a][b] = mp[b][a] = x;
}
scanf("%d%d", &s, &t);
Floyd();
if (mp[s][t] != INF) {
printf("%d\n", mp[s][t]);
}
else {
printf("-1\n");
}
}
return 0;
}