HDU 6005 Pandaland(无向图最小环)

题意：

给出无向图，然后求最小环，点较多，边<=4000。

思路：

求最小环，对于n较少的题floyd改版来解决，而这题点很多而边不是很多，所以只能通过枚举每一条边，然后再求去掉这条边之后，其两个端点之间的最短路，从而来寻找最小环，时间复杂度为O(m*m*logn)。

代码：

#include <bits/stdc++.h>#define ll long longusing namespace std;const ll inf = 0x3f3f3f3f3f3f3f3f;const int maxn = 8005;const int bas = 20001;struct node {int v; ll w;bool operator<(const node k)const {return w > k.w;}};priority_queue<node> q;struct node1 {ll w; int u, v, next;} edge[maxn];int no, head[maxn];ll dis[maxn];bool col[maxn], vis[maxn];unordered_map<int, int> has;int t, m, n;void init(){no = 0;memset(head, -1, sizeof head);}inline void add(int u, int v, ll w){edge[no].u = u;edge[no].v = v;edge[no].w = w;edge[no].next = head[u];head[u] = no++;}ll DJ(int S, int T){memset(vis, 0, sizeof vis);while(!q.empty()) q.pop();memset(dis, 0x3f3f, sizeof dis);dis[S] = 0;q.push((node){S, 0});while(!q.empty()){node tp = q.top(); q.pop();int u = tp.v;if(u == T) return dis[T];if(vis[u]) continue;vis[u] = 1;for(int k = head[u]; k+1; k = edge[k].next){if(col[k]) continue;int v = edge[k].v;if(dis[v] > dis[u]+edge[k].w){dis[v] = dis[u]+edge[k].w;q.push((node){v, dis[v]});}}}return inf;}void work(){ll ans = inf;memset(col, 0, sizeof col);for(int i = 0; i < no; i+=2){int u = edge[i].u, v = edge[i].v;col[i] = col[i+1] = 1;ll _dis = DJ(u, v);if(_dis+edge[i].w < ans) ans = _dis+edge[i].w;col[i] = col[i+1] = 0;}if(ans == inf) puts("0");else printf("%lld\n", ans);}int main(){scanf("%d", &t);for(int _ = 1; _ <= t; ++_){init();n = 0; has.clear();scanf("%d", &m);for(int i = 1; i <= m; ++i){int _x1, _y1, _x2, _y2, w, u, v;scanf("%d %d %d %d %d", &_x1, &_y1, &_x2, &_y2, &w);_x1 += 10000, _y1 += 10000, _x2 += 10000, _y2 += 10000;if(has.find(_x1*bas+_y1) == has.end())has[_x1*bas+_y1] = u = ++n;elseu = has[_x1*bas+_y1];if(has.find(_x2*bas+_y2) == has.end())has[_x2*bas+_y2] = v = ++n;elsev = has[_x2*bas+_y2];add(u, v, w);add(v, u, w);}printf("Case #%d: ", _);work();}return 0;}

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