HDU 6073 Matching In Multiplication(拓扑排序)
Matching In Multiplication
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1127 Accepted Submission(s): 325
Little Q misunderstands the definition of bipartite graph, he thinks the size of U is equal to the size of V, and for each vertex p in U, there are exactly two edges from p. Based on such weighted graph, he defines the weight of a perfect matching as the product of all the edges' weight, and the weight of a graph is the sum of all the perfect matchings' weight.
Please write a program to compute the weight of a weighted ''bipartite graph'' made by Little Q.
In each test case, there is an integer n(1≤n≤300000) in the first line, denoting the size of U. The vertex in U and V are labeled by 1,2,...,n.
For the next n lines, each line contains 4 integers vi,1,wi,1,vi,2,wi,2(1≤vi,j≤n,1≤wi,j≤109), denoting there is an edge between Ui and Vvi,1, weighted wi,1, and there is another edge between Ui and Vvi,2, weighted wi,2.
It is guaranteed that each graph has at least one perfect matchings, and there are at most one edge between every pair of vertex.
2
2 1 1 4
1 4 2 3
#include <bits/stdc++.h>
#define inf 0x3f3f3f3f
#define met(a,b) memset(a,b,sizeof a)
#define pb push_back
#define mp make_pair
#define inf 0x3f3f3f3f
#define qwer 2e18
using namespace std;
typedef long long ll;
const int N = 6e5+;
const int M = ;
const int mod = ;
const double pi= acos(-1.0);
typedef pair<int,int>pii;
int n,s;
int vis[N],in[N];
ll ans[];
vector<pii>edg[N];
ll topSort(){
queue<int>q;
ll ret=;
for(int i=n+;i<=n+n;i++){
if(in[i]==){
q.push(i);
vis[i]=;
}
}
while(!q.empty()){
int u=q.front();
q.pop();
for(int i=;i<edg[u].size();i++){
int v=edg[u][i].first;
if(vis[v])continue;
if((--in[v])==)q.push(v),vis[v]=;
if(u>n)ret=(ret*1LL*edg[u][i].second)%mod;
}
}
return ret;
}
void dfs(int u,int ty,int fa){
vis[u]=;
for(int i=;i<edg[u].size();i++){
int v=edg[u][i].first;
if(v==s&&v!=fa)ans[ty]=(ans[ty]*1LL*edg[u][i].second)%mod;
if(vis[v])continue;
ans[ty]=(ans[ty]*1LL*edg[u][i].second)%mod;
dfs(v,ty^,u);
}
}
int main(){
//freopen("de.txt","r",stdin);
int T;
scanf("%d",&T);
while(T--){
scanf("%d",&n);
for(int i=;i<N;i++)vis[i]=in[i]=,edg[i].clear();
for(int i=,v1,w1,v2,w2;i<=n;i++){
scanf("%d%d%d%d",&v1,&w1,&v2,&w2);
v1+=n;v2+=n;
edg[i].pb(mp(v1,w1));
edg[v1].pb(mp(i,w1));
edg[i].pb(mp(v2,w2));
edg[v2].pb(mp(i,w2));
in[i]+=;
in[v1]++;in[v2]++;
}
ll anss=topSort();
for(s=;s<=n;s++){
if(!vis[s]){
ans[]=ans[]=;
dfs(s,,);
anss=anss*((ans[]+ans[])%mod)%mod;
}
}
printf("%lld\n",anss);
}
return ;
}
HDU 6073 Matching In Multiplication(拓扑排序)的更多相关文章
- HDU 6073 - Matching In Multiplication | 2017 Multi-University Training Contest 4
/* HDU 6073 - Matching In Multiplication [ 图论 ] | 2017 Multi-University Training Contest 4 题意: 定义一张二 ...
- HDU 6073 Matching In Multiplication(拓扑排序+思维)
http://acm.hdu.edu.cn/showproblem.php?pid=6073 题意:有个二分图,左边和右边的顶点数相同,左边的顶点每个顶点度数为2.现在有个屌丝理解错了最佳完美匹配,它 ...
- HDU 6073 Matching In Multiplication dfs遍历环 + 拓扑
Matching In Multiplication Problem DescriptionIn the mathematical discipline of graph theory, a bipa ...
- HDU 6073 Matching In Multiplication —— 2017 Multi-University Training 4
Matching In Multiplication Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K ( ...
- 2017 ACM暑期多校联合训练 - Team 4 1007 HDU 6073 Matching In Multiplication (模拟)
题目链接 Problem Description In the mathematical discipline of graph theory, a bipartite graph is a grap ...
- HDU.3342 Legal or Not (拓扑排序 TopSort)
HDU.3342 Legal or Not (拓扑排序 TopSort) 题意分析 裸的拓扑排序 根据是否成环来判断是否合法 详解请移步 算法学习 拓扑排序(TopSort) 代码总览 #includ ...
- HDU.1285 确定比赛名次 (拓扑排序 TopSort)
HDU.1285 确定比赛名次 (拓扑排序 TopSort) 题意分析 裸的拓扑排序 详解请移步 算法学习 拓扑排序(TopSort) 只不过这道的额外要求是,输出字典序最小的那组解.那么解决方案就是 ...
- HDU 4857 逃生 (反向拓扑排序 & 容器实现)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4857 逃生 Time Limit: 2000/1000 MS (Java/Others) Mem ...
- ACM: HDU 1285 确定比赛名次 - 拓扑排序
HDU 1285 确定比赛名次 Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u De ...
随机推荐
- python测试rabbitmq简易实例
生产者 import pika #coding=utf8 credentials = pika.PlainCredentials('guest', '密码') connection = pika.Bl ...
- java -classpath or -cp 的设置和解释
classpath is a parameter—set either on the command-line, or through an environment variable—that tel ...
- Docker explainations
What does docker run --link mean, what's the usage? link 是在两个contain之间建立一种父子关系,父container中的web,可以得到子 ...
- JS 控制页面刷新
.页面自动刷新:把如下代码加入<head>区域中 <meta http-equiv=">,其中20指每隔20秒刷新一次页面. .页面自动跳转:把如下代码加入<h ...
- io多路复用-select()
参照<Unix网络编程>相关章节内容,实现了一个简单的单线程IO多路复用服务器与客户端. 普通迭代服务器,由于执行recvfrom则会发生阻塞,直到客户端发送数据并正确接收后才能够返回,一 ...
- device tree source file position
android/kernel/msm-4.9/arch/arm64/boot/dts/qcom/
- python基础===tkinter学习链接
http://effbot.org/tkinterbook/tkinter-classes.htm
- linux加载指定目录的so文件
linux加载指定目录的so文件 http://blog.csdn.net/win_lin/article/details/8286125 download urlhttp://download.ch ...
- Deep Learning基础--线性解码器、卷积、池化
本文主要是学习下Linear Decoder已经在大图片中经常采用的技术convolution和pooling,分别参考网页http://deeplearning.stanford.edu/wiki/ ...
- VPS性能综合测试(7):服务器压力测试,VPS系统负载测试
1.可能有的VPS主机使用性能测评工具得出的结果很优秀,但是最终运用到实际生产时却发现VPS主机根本无法承受理论上应该达到的流量压力,这时我们就不得不要怀疑VPS商是不是对VPS主机的参数进行了“篡改 ...