Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)

Total Submission(s): 6987    Accepted Submission(s): 3262

Problem Description
Network flow is a well-known difficult problem for ACMers. Given a graph, your task is to find out the maximum flow for the weighted directed graph.
 
Input
The first line of input contains an integer T, denoting the number of test cases.

For each test case, the first line contains two integers N and M, denoting the number of vertexes and edges in the graph. (2 <= N <= 15, 0 <= M <= 1000)

Next M lines, each line contains three integers X, Y and C, there is an edge from X to Y and the capacity of it is C. (1 <= X, Y <= N, 1 <= C <= 1000)
 
Output
For each test cases, you should output the maximum flow from source 1 to sink N.
 
Sample Input
2
3 2
1 2 1
2 3 1
3 3
1 2 1
2 3 1
1 3 1
 
Sample Output
Case 1: 1
Case 2: 2
 
Author
HyperHexagon

解题报告

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define N 30
#define inf 99999999
using namespace std;
int n,m,a[N],flow,pre[N];
int Edge[N][N];
queue<int >Q;
int bfs()
{
while(!Q.empty())
Q.pop();
memset(a,0,sizeof(a));
memset(pre,0,sizeof(pre));
Q.push(1);
pre[1]=1;
a[1]=inf;
while(!Q.empty())
{
int u=Q.front();
Q.pop();
for(int v=1;v<=n;v++)
{
if(!a[v]&&Edge[u][v]>0)
{
pre[v]=u;
a[v]=min(Edge[u][v],a[u]);
Q.push(v);
}
}
if(a[n])break;
}
if(!a[n])
return -1;
else return a[n];
}
void ek()
{
int a,i;
while((a=bfs())!=-1)
{
for(i=n;i!=1;i=pre[i])
{
Edge[pre[i]][i]-=a;
Edge[i][pre[i]]+=a;
}
flow+=a;
}
}
int main()
{
int t,i,j,u,v,w,k=1;
scanf("%d",&t);
while(t--)
{
flow=0;
memset(Edge,0,sizeof(Edge));
scanf("%d%d",&n,&m);
for(i=0;i<m;i++)
{
scanf("%d%d%d",&u,&v,&w);
Edge[u][v]+=w;
}
ek();
printf("Case %d: ",k++);
printf("%d\n",flow);
}
return 0;
}

HDU3549_Flow Problem(网络流/EK)的更多相关文章

  1. HDU 3549 Flow Problem 网络流(最大流) FF EK

    Flow Problem Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others) Tot ...

  2. HDU 3549 基础网络流EK算法 Flow Problem

    欢迎参加——BestCoder周年纪念赛(高质量题目+多重奖励) Flow Problem Time Limit: 5000/5000 MS (Java/Others)    Memory Limit ...

  3. 网络流EK

    #include <iostream> #include <queue> #include <string.h> #define MAX 302 using nam ...

  4. HDU1532 Drainage Ditches 网络流EK算法

    Drainage Ditches Problem Description Every time it rains on Farmer John's fields, a pond forms over ...

  5. hdu 3549 Flow Problem 网络流

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549 Network flow is a well-known difficult problem f ...

  6. POJ-3436 ACM Computer Factory(网络流EK)

    As you know, all the computers used for ACM contests must be identical, so the participants compete ...

  7. 网络流Ek算法

    例题:  Flow Problem HDU - 3549 Edmonds_Karp算法其实是不断找增广路的过程. 但是在找的过程中是找"最近"的一天增广路, 而不是找最高效的一条增 ...

  8. 【HDU5772】String Problem [网络流]

    String Problem Time Limit: 10 Sec  Memory Limit: 64 MB[Submit][Status][Discuss] Description Input Ou ...

  9. [bzoj3218]a + b Problem 网络流+主席树优化建图

    3218: a + b Problem Time Limit: 20 Sec  Memory Limit: 40 MBSubmit: 2229  Solved: 836[Submit][Status] ...

随机推荐

  1. border:none;和border:0;的区别

    一.是理论上的性能差异 [border:0;]把border设为“0”像素虽然在页面上看不见,但按border默认值理解,浏览器依然对border-width/border-color进行了渲染,即已 ...

  2. 关于platform_device和platform_driver的匹配【转】

    转自:http://blog.csdn.net/dfysy/article/details/5959451 版权声明:本文为博主原创文章,未经博主允许不得转载. 说句老实话,我不太喜欢现在Linux ...

  3. 《手把手教你学C语言》学习笔记(4)---代码规范

    编程过程中需要遵守编译器的各种约定,例如以下代码: 1 #include <stdio.h> 2 3 int main(int argc, char **argv) 4 { 5 print ...

  4. Python Challenge 第十五关

    第15关,题目是 whom? 有一张图片,是个日历.日历的年份是 1XX6,中间是被挖去的洞.然后图中1月26日被画了个圈,当天是星期一.右下角的二月小图中有29号,可以得知这是闰年.然后查看源代码. ...

  5. ural 1519 fomular 1 既插头DP学习笔记

    直接看CDQ在2008年的论文吧. 个人认为她的论文有两个不明确的地方, 这里补充一下: 首先是轮廓的概念. 我们在进行插头DP时, 是从上往下, 从左往右逐个格子进行的, 已经处理的格子与未经处理的 ...

  6. maxwell简单部署使用

    详细资料可以参考maxwell官网  (mysql + maxwell + kafka + elasticsearch) 说明:本文主要是关于配置maxwell监听mysql的数据修改并实时将修改内容 ...

  7. 以root用户身份在jenkins中运行shell命令

    以下过程是CentOS 1.打开此脚本(使用VIM或其他编辑器): vim /etc/sysconfig/jenkins 2.找到$JENKINS_USER并更改为“root”: $JENKINS_U ...

  8. ASIHTTPRequest实现断点续传

    http://blog.csdn.net/daiyelang/article/category/1377418 ASIHTTPRequest可以实现断点续传.网上有一些介绍类似使用:   [reque ...

  9. MFC MFC对话框滚动条的使用

      对话框的(上下/左右)滚动事件,比如,把一个比较大的对话框放入tab控件的某一页时,就需要添加滚动条.在使用了java和qt等图形界面化的集成开发环境之后,再使用MFC,就会发现,想要让一个对话框 ...

  10. 如何从底层调试docker

    How the docker container creation process works (from docker run to runc) Over the past few months I ...