floyd求最短路

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
using namespace std;
inline char gc() { 
    static char buf[1 << 18], *fs, *ft;
    return (fs == ft && (ft = (fs = buf) + fread(buf, 1, 1 << 18, stdin)), fs == ft) ? EOF : *fs++;
}
inline int read(){
    int num=0,f=1; char c=getchar();
    while(!isdigit(c)){if(c==‘-‘) f=-1; c=getchar();}
    while(isdigit(c)){num=(num<<1)+(num<<3)+(c^48); c=getchar();}
    return num*f;
}
#define maxn 2000
int dis[maxn][maxn],n,m;
inline int in(){
    memset(dis,10,sizeof(dis));
    n=read(); m=read();
    for(int i=1;i<=m;i++){
        int u,v,c;
        u=read(); v=read(); c=read();
        dis[u][v]=c;
    } 
}
inline int floyd(){
    for(int k=1;k<=n;k++)
       for(int i=1;i<=n;i++)
          for(int j=1;j<=n;j++)
             dis[i][j]=min(dis[i][j],dis[i][k]+dis[k][j]);
}
inline int out(){
    cout<<dis[1][n];
}
int main(){
    in();
    floyd();
    out();
}

floyd可以处理多源最短路问题