单源最短路径<Dijkstra>

由一个源头，也就是n1开始到n2。

f[?]代表从终点到？的距离，刚开始f[n2]就是确定了终点，然后每个f[?]为距离

需要弄明白map的判断那里

#include <iostream>
#include <string.h>
#define maxn 9999
using namespace std;
int main(){
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
int n,n1,n2,x;
cin >> n >> n1 >> n2; //n表示有n个点，从n1到n2的距离
int map[n+1][n+1];
int f[n+1],c[n+1];
int t1=1,t2=1,t3=1,i,j;
for(i=1;i<=n;i++)
for(j=1;j<=n;j++)
map[i][j]=-1;
while(t1&&t2&&t3){
cin >> t1 >> t2 >> t3;
map[t1][t2]=t3;
}
for(i=1;i<=n;i++)
f[i]=maxn;
f[n2]=0;
for(i=n;i>0;i--)
for(j=1;j<=n;j++)
if(map[i][j]!=-1&&f[j]!=maxn&&f[j]+map[i][j]<f[i]){
f[i]=f[j]+map[i][j];
//c[i]=j;
}
cout << f[n1]<<endl;
/*x=n1;
while(x){
printf("%d",x);
x=c[x];
}*/return 0;
}