description:
The two integers n and m in the first line represent the number of vertices (numbered from 1 to n), and m represents the number of edges. In the next m rows, each row has 3 numbers x, y, z, which express the weight of the edge from vertex x to vertex y as z.
Sample
enter:
6 9
1 2 1
1 3 12
2 3 9
2 4 3
3 5 5
4 3 4
4 5 13
4 6 15
5 6 4
Output:0 1 8 4 13 17
Code:
#include<bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f; //定义无穷大
int main(){
//freopen("a.txt","r",stdin);
int n,m,e[200][200],d[200],vis[200],min;
//n为顶点个数，m为边的数目，e为邻接矩阵，d为最短路数组，vis标记节点是否访问过，min为最小值
cin>>n>>m;
int u,v,k;
memset(d,INF,sizeof d);
memset(e,INF,sizeof e);
memset(vis,0,sizeof vis);
for(int i = 1;i<=m;i++){
cin>>u>>v>>k;
e[u][v] = k;
}
//读入边
//以下为核心：
//一、 初始化，将出发点到自身的距离设置为0
d[1] = 0;
//二、循环n（顶点个数）次
for(int i =1;i<=n;i++){
//三、将最小值设置为无穷大，方便寻找最小值
min = INF;
//四、循环n次，每次都寻找当前未访问过顶点中的路径最小的顶点
for(int j = 1;j<=n;j++){
if(vis[j] == 0&& d[j] < min){
min = d[j];
u = j;
}
}
//五、将寻找到的当前路径最小点设置为已经访问
vis[u] = 1;
//六、判断是否可以优化
for(v = 1;v<= n;v++){
if(vis[v] ==0 && e[u][v]<INF ){
if( d[u] + e[u][v] < d[v]){
d[v] = d[u] + e[u][v];
}
}
}
}
//输出
for(int i = 1;i<=n;i++){
cout<<d[i]<<' ';
}
return 0;
}
end.