hdu 6073 Matching In Multiplication(2017 Multi-University Training Contest - Team 4 )
2017-08-05 09:11
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Matching In Multiplication
题目链接:Matching In Multiplication题意:给你一个二分图,集合U和V各有n个点,集合U的每个点都连出两条边。保证至少有一个完美匹配。对于一个完美匹配,价值是边权之积,要求所有完美匹配的价值和。
官方题解:
思路:
如果一个点的度数为1的话,那么它的匹配方案肯定是固定的,因此我们可以先通过拓扑排序去掉集合V中度数为1的点,对V中度数为1的点都在U中找一个点与之匹配。那么这些对答案ans的贡献为ans*唯一对应边的边权
假设集合U和V中现在都只剩下k个点,由于集合U中点的度数都为2,集合V中点的度数都≥2,所以集合V的每个点的度数肯定也都为2。因此现在图中的每个连通块肯定都成环。
那么我们只需要对每个连通块间隔取边权,每个连通块的完备匹配权值为 part[0] part[1] 。则该连通块对 ans 的贡献为 ans×(part0+part1) 。
代码:
#include<bits/stdc++.h> using namespace std; typedef long long LL; const int maxn=6e5+10; const int mod=998244353; struct edge { int v,w,next,mark; } E[maxn<<1]; int first[maxn],vis[maxn],deg[maxn]; LL part[3]; int n,len; void add_edge(int u,int v,int w) { E[len].v=v,E[len].w=w,E[len].mark=0,E[len].next=first[u],first[u]=len++; } void dfs(int u,int idx) { vis[u]=1; for(int i=first[u]; ~i; i=E[i].next) { if(E[i].mark) continue; E[i].mark=E[i^1].mark=1; part[idx]=part[idx]*E[i].w%mod; dfs(E[i].v,idx^1); } } int main() { int T_T; scanf("%d",&T_T); while(T_T--) { memset(deg,0,sizeof(deg)); memset(vis,0,sizeof(vis)); memset(first,-1,sizeof(first)); scanf("%d",&n); len=0; for(int u=1,v,w; u<=n; ++u) { for(int i=1; i<=2; ++i) { scanf("%d%d",&v,&w); add_edge(u,v+n,w); add_edge(v+n,u,w); ++deg[u],++deg[v+n]; } } LL ans=1; queue<int>q; for(int v=n+1; v<=n+n; ++v) if(deg[v]==1) q.push(v); while(!q.empty()) { int v=q.front(); q.pop(),vis[v]=1; for(int i=first[v]; ~i; i=E[i].next) { if(E[i].mark) continue; E[i].mark=E[i^1].mark=1; vis[E[i].v]=1; ans=(ans*E[i].w)%mod; for(int j=first[E[i].v]; ~j; j=E[j].next) { E[j].mark=E[j^1].mark=1; --deg[E[j].v]; if(deg[E[j].v]==1) q.push(E[j].v); } } } for(int i=1; i<=n; ++i) { if(vis[i]) continue; part[0]=part[1]=1; dfs(i,0); ans=ans*((part[0]+part[1])%mod)%mod; } printf("%lld\n",ans); } return 0; }
参考博客:
DorMOUSENone
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