short-path problem (Dijkstra) 分类: ACM TYPE 2014-09-01 23:51 111人阅读 评论(0) 收藏

#include <cstdio>
#include <iostream>
#include <cstring> using namespace std;
const int INF = 0x3fffffff; int g[1005][1005];
int m; int Dijkstra(int s,int t)
{
bool visit[1005];
int dis[1005];
for(int i = 1; i <= m; ++i)
{
visit[i] = false;
dis[i] = g[s][i];
}
visit[s] = true; int min = INF;
int k;
for(int i = 1; i < m; ++i)
{
min = INF;
k = 1;
for(int j = 1; j <= m; ++j)
{
if(!visit[j] && dis[j] < min)
{
min = dis[j];
k = j;
}
}
visit[k] = true;
for(int j = 1; j <= m; ++j)
{
if(!visit[j] && dis[k] + g[k][j] < dis[j])
{
dis[j] = dis[k] + g[k][j];
}
}
}
return dis[t];
} int main()
{
int T, a, b, len;
int s, t, p;
scanf("%d", &T);
while(T--)
{
scanf("%d%d", &m, &p);
scanf("%d%d",&s,&t);
for(int i = 1; i <= m; ++i)
for(int j = 1; j < i; ++j)
g[i][j] = g[j][i] = INF; while(p--)
{
scanf("%d%d%d", &a, &b, &len);
if(g[a][b] > len)
g[a][b] = g[b][a] = len;
} printf("%d\n", Dijkstra(s,t));
}
return 0;
}

Code From Nyoj 115

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