HDU 1599 find the mincost route

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把每一条边删除掉然后求一次这条边顶点之间的最短距离,最短距离加上删除掉的边的权值就是环的权值,然后求出最小环值.

如果删除了这条边,两点不连通了说明两点无法构成一个环.

m次dijkstra

#include <iostream>#include <cstdio>#include <memory.h>#include <vector>#include <queue>#include <functional>#define inf 0X20202020#define pii pair<int,int>using namespace std;const int maxn=105;int g[maxn][maxn],dis[maxn],N,M;bool vis[maxn];int ans;int dijkstra(int s,int e){priority_queue<pii,vector<pii>,greater<pii> >q;memset(vis,0,sizeof(vis));memset(dis,0X20,sizeof(dis));q.push(make_pair(0,s));dis[s]=0;while (q.size()){pii t=q.top();q.pop();if(t.second==e)return dis[e];else if(vis[t.second])continue;for (int i=1;i<=N;++i){if(g[t.second][i]!=inf&&dis[i]>dis[t.second]+g[t.second][i]&&!vis[i]){dis[i]=dis[t.second]+g[t.second][i];q.push(make_pair(dis[i],i));}}}return inf;}int main(){while (scanf("%d%d",&N,&M)==2){memset(g,0X20,sizeof(g));for (int i=0;i<M;++i){int u,v,w;scanf("%d%d%d",&u,&v,&w);g[u][v]=g[v][u]=min(g[u][v],w);}ans=inf;for (int i=1;i<=N;++i){for (int j=i+1;j<=N;++j){if(g[i][j]!=inf){int t=g[i][j];g[i][j]=g[j][i]=inf;//删除int cost=t+dijkstra(i,j);g[i][j]=g[j][i]=t;//还原ans=min(ans,cost);}}}if(ans==inf){printf("It's impossible.\n");}else{printf("%d\n",ans);}}return 0;}