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ime Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 32657 Accepted Submission(s): 11929
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define INF 1<<30
using namespace std;
int n,m;
int st,ed;
int w[205][205];
int d[205],vis[205];
void read_graph()//邻接矩阵存储图
{
for(int i=0;i<205;++i)
for(int j=0;j<205;++j)
w[i][j]=INF;
int u,v,c;
for(int i=0;i<m;++i){
cin>>u>>v>>c;
if(c>w[u][v]) continue;//WA了十几次,最终看了DIC才知道相同路径的权值可能不同,应保存较小的
w[u][v]=c;//如 2 3 2和2 3 4 后一个应舍弃
w[v][u]=c;
}
cin>>st>>ed;
}
void dij()
{
memset(vis,0,sizeof(vis));
for(int i=0;i<=n;++i) d[i]=INF;
d[st]=0;
for(int i=0;i<n;++i){
int x,m=INF;
for(int j=0;j<n;++j) if(!vis[j]&&d[j]<=m) m=d[x=j];
vis[x]=1;
for(int j=0;j<n;++j)
if(w[x][j]!=INF) d[j]=min(d[j],d[x]+w[x][j]);
}
}
void solve()
{
read_graph();
dij();
//for(int i=0;i<n;++i) cout<<d[i]<<endl;
if(d[ed]==INF) cout<<-1<<endl;//若起点到终点的最短路仍为初始值INF,则不通
else cout<<d[ed]<<endl;
}
int main()
{
//freopen("case.txt","r",stdin);
while(~scanf("%d%d",&n,&m))
solve();
return 0;
}
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原文地址:http://www.cnblogs.com/orchidzjl/p/4409303.html