hdu6005 Pandaland 想法+dijkstra

Posted 2020-09-30 天道酬勤007

/**
题目：hdu6005 Pandaland
链接：http://acm.hdu.edu.cn/showproblem.php?pid=6005
题意：给定一个带权无向图，求权值和最小的环的值，如果不存在环输出0；

思路：枚举每条边，然后dijkstra求s到t的距离，dijkstra过程中舍去s-t的这条边。
两个优化：dijkstra找到了t就跳出。或者出队列的距离>=当前找到的最小距离跳出。


*/

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<map>
#include<vector>
#include<queue>
#include<cstring>
using namespace std;
typedef pair<int,int> P;
typedef long long LL;
const int N = 4e3+10;
const int INF = 0x3f3f3f3f;
struct edge{int to, cost;};
int V, mis;
vector<edge> G[2*N];
int d[N*2+1];
void dijkstra(int s,int t)
{
    priority_queue<P,vector<P>, greater<P> > que;
    fill(d,d+V,INF);
    d[s] = 0;
    que.push(P(0,s));
    while(!que.empty()){
        P p = que.top(); que.pop();
        int v = p.second;
        if(d[v]<p.first) continue;
        if(v==t) break;
        if(d[v]>=mis) break;
        for(int i = 0; i < (int)G[v].size(); i++){
            edge e = G[v][i];
            if(v==s&&e.to==t) continue;
            if(d[e.to]>d[v]+e.cost){
                d[e.to] = d[v]+e.cost;
                que.push(P(d[e.to],e.to));
            }
        }
    }
}
map<P,int> mp;
struct node
{
    int u,v;
    int cost;
}Eg[N];
int main()
{
    int cas = 1, T, m;
    cin>>T;
    while(T--)
    {
        int cnt = 1;
        scanf("%d",&m);
        int cost, x1, y1, x2, y2;
        for(int i = 1; i <= 2*N; i++) G[i].clear();
        mp.clear();
        for(int i = 1; i <= m; i++){
            scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&cost);
            int u, v;
            if(mp[P(x1,y1)]==0){
                mp[P(x1,y1)] = cnt++;
            }
            if(mp[P(x2,y2)]==0){
                mp[P(x2,y2)] = cnt++;
            }
            u = mp[P(x1,y1)], v = mp[P(x2,y2)];
            G[u].push_back(edge{v,cost});
            G[v].push_back(edge{u,cost});
            Eg[i].cost = cost;
            Eg[i].u = u;
            Eg[i].v = v;
        }
        mis = INF;
        V = cnt;
        for(int i = 1; i <= m; i++){
            dijkstra(Eg[i].u,Eg[i].v);
            mis = min(mis,d[Eg[i].v]+Eg[i].cost);
        }
        if(mis==INF){
            printf("Case #%d: 0\n",cas++);
        }else
        printf("Case #%d: %d\n",cas++,mis);
    }
    return 0;
}