Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 11405    Accepted Submission(s):
5418

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
 
 
刚开始看不是太理解  解析会后续更新
#include<stdio.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<algorithm>
#define INF 0x7fffff
#define MAX 2100
using namespace std;
int ans,head[MAX];
int n,m;
int dis[MAX],vis[MAX];
int cur[MAX];
struct node
{
int beg,end,cap,flow,next;
}edge[MAX];
void init()
{
ans=0;
memset(head,-1,sizeof(head));
}
void add(int u,int v,int w)
{
edge[ans].beg=u;
edge[ans].end=v;
edge[ans].cap=w;
edge[ans].flow=0;
edge[ans].next=head[u];
head[u]=ans++;
}
void getmap()
{
int i,a,b,c;
while(m--)
{
scanf("%d%d%d",&a,&b,&c);
add(a,b,c);
add(b,a,0);
}
}
int bfs(int start,int over)
{
int i,v;
memset(dis,-1,sizeof(dis));
memset(vis,0,sizeof(vis));
queue<int>q;
while(!q.empty())
q.pop();
q.push(start);
vis[start]=1;
dis[start]=0;
while(!q.empty())
{
int u=q.front();
q.pop();
for(i=head[u];i!=-1;i=edge[i].next)
{
v=edge[i].end;
if(!vis[v]&&edge[i].cap>edge[i].flow)
{
vis[v]=1;
dis[v]=dis[u]+1;
if(v==over)
return 1;
q.push(v);
}
}
}
return 0;
}
int dfs(int x,int a,int over)
{
if(x==over||a==0)
return a;
int flow=0,f;
for(int& i=cur[x];i!=-1;i=edge[i].next)
{
if(dis[x]+1==dis[edge[i].end]&&(f=dfs(edge[i].end,min(a,edge[i].cap-edge[i].flow),over))>0)
{
edge[i].flow+=f;
edge[i^1].flow-=f;
flow+=f;
a-=f;
if(a==0) break;
}
}
return flow;
}
int maxflow(int start,int over)
{
int flow=0;
while(bfs(start,over))
{
memcpy(cur,head,sizeof(head));
flow+=dfs(start,INF,over);
}
return flow;
}
int main()
{
int t,k,j,i;
scanf("%d",&t);
k=1;
while(t--)
{
scanf("%d%d",&n,&m);
init();
getmap();
printf("Case %d: ",k++);
printf("%d\n",maxflow(1,n));
}
return 0;
}

  

 

hdoj 3549 Flow Problem【网络流最大流入门】的更多相关文章

  1. 题解报告:hdu 3549 Flow Problem(最大流入门)

    Problem Description Network flow is a well-known difficult problem for ACMers. Given a graph, your t ...

  2. HDU 3549 Flow Problem(最大流)

    HDU 3549 Flow Problem(最大流) Time Limit: 5000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/ ...

  3. hdu 3549 Flow Problem【最大流增广路入门模板题】

    题目:http://acm.hdu.edu.cn/showproblem.php?pid=3549 Flow Problem Time Limit: 5000/5000 MS (Java/Others ...

  4. 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 ...

  5. HDU 3549 Flow Problem(最大流模板)

    http://acm.hdu.edu.cn/showproblem.php?pid=3549 刚接触网络流,感觉有点难啊,只好先拿几道基础的模板题来练练手. 最大流的模板题. #include< ...

  6. HDU 3549 Flow Problem (最大流ISAP)

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

  7. hdu 3549 Flow Problem (网络最大流)

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

  8. hdoj 3549 Flow Problem(最大网络流)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549 思路分析:该问题为裸的最大网络流问题,数据量不大,使用EdmondsKarp算法求解即可:需要注 ...

  9. 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 ...

随机推荐

  1. Automotive Security的一些资料和心得(6):AUTOSAR

    1.1 Introduction AUTOSAR(汽车开放系统架构)是一个开放的,标准化的汽车软件架构,由汽车制造商,供应商和开发工具共同开发.它联合了汽车OEM ,供应商和开发工具供应商,其目标是创 ...

  2. jquery delegate

    代码如下:   $('#container').delegate('a','click',function(){alert('That tickles!')} JQuery扫描文档查找$('#cont ...

  3. yum和rpm命令详解

    rpm,全称RPM Package Manager,是RedHat发布的,针对特定硬件,已经编译好的软件包.安装之后就可以使用,不需要自行编译,以及之前对软件和硬件的检测,目录的配置等动作. yum, ...

  4. sjtu1590 强迫症

    Description BS96发布了一套有\(m\)个band柄绘的新badge,kuma先生想要拿到04的badge于是进行了抽抽抽. kuma先生一共抽了\(n\)个badge.他把所有的bad ...

  5. HDU4523+简单

    题意很简单. 一次最多多切出一条边! 其余的就没什么好说的了 import java.util.*; import java.math.*; public class Main{ public sta ...

  6. IDM和ODM

    DM (Integrated Data Multiplexer):综合数据复用器[1]  综合数据复用器是一种数据复用设备,它可以将多路RS232.RS485及数字语音等多种数据复用到E1传输通道或光 ...

  7. Android handler 报错处理Can't create handler inside thread that has not called Looper.prepare()

    问题: 写了一个sdk给其他人用,提供一个回调函数,函数使用了handler处理消息 // handler监听网络请求,完成后操作回调函数 final Handler trigerGfHandler ...

  8. Which are in?

    Which are in? Given two arrays of strings a1 and a2 return a sorted array in lexicographical order a ...

  9. Linux -- Ubuntu搭建java开发环境

    Steps 1 Check to see if your Ubuntu Linux operating system architecture is 32-bit or 64-bit, open up ...

  10. 下拉列表联动显示(Car表) 三级联动

    .Models namespace 下拉列表联动显示_Car表_.Models { public class ProductorBF { private MyDBDataContext _contex ...