cdoj 30 最短路 flyod

题目链接:

http://acm.uestc.edu.cn/#/problem/show/30
题意:

题解:

直接floyd
代码:

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 #define MS(a) memset(a,0,sizeof(a))
 5 #define MP make_pair
 6 #define PB push_back
 7 const int INF = 0x3f3f3f3f;
 8 const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
 9 inline ll read(){
10     ll x=0,f=1;char ch=getchar();
11     while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
12     while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
13     return x*f;
14 }
15 //////////////////////////////////////////////////////////////////////////
16 const int maxn = 100+10;
17 
18 int d[maxn][maxn];
19 
20 int main(){
21     int n,m;
22     while(scanf("%d%d",&n,&m),n+m){
23         for(int i=0; i<=n; i++)
24             for(int j=0; j<=n; j++)
25                 d[i][j] = INF;
26         for(int i=0; i<m; i++){
27             int u,v,w; scanf("%d%d%d",&u,&v,&w);
28             d[u][v] = d[v][u] = w;
29         }
30 
31         for(int k=1; k<=n; k++)
32             for(int i=1; i<=n; i++)
33                 for(int j=1; j<=n; j++)
34                     d[i][j] = min(d[i][j],d[i][k]+d[k][j]);
35 
36         cout << d[1][n] << endl;
37     }
38 
39     return 0;
40 }