Flow Problem

Time Limit: 5000/5000 MS (Java/Others)    Memory Limit: 65535/32768 K (Java/Others)
Total Submission(s): 8199    Accepted Submission(s): 3814

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
 
题意:给一个有向图,求源点为1汇点为n的最大流
 
思路:裸题,ISAP。。原来之前自己写的模版有点疏漏了。。
 
#include <vector>
#include <cstdio>
#include <cstring>
#include <queue>
#define FOR(i,n) for(i=1;i<=(n);i++)
using namespace std;
const int INF = 1e9;
const int N = ; struct Edge{
int from,to,cap,flow;
}; struct ISAP{
int n,m,s,t;
int p[N],num[N];
vector<Edge> edges;
vector<int> G[N];
bool vis[N];
int d[N],cur[N];
void init(int _n,int _m)
{
n=_n; m=_m;
int i;
edges.clear();
FOR(i,n)
{
G[i].clear();
d[i]=INF;
}
}
void AddEdge(int from,int to,int cap)
{
edges.push_back((Edge){from,to,cap,});
edges.push_back((Edge){to,from,,});
m = edges.size();
G[from].push_back(m-);
G[to].push_back(m-);
}
bool BFS()
{
memset(vis,,sizeof(vis));
queue<int> Q;
Q.push(t);
d[t]=;
vis[t]=;
while(!Q.empty())
{
int x = Q.front(); Q.pop();
for(unsigned i=;i<G[x].size();i++)
{
Edge& e = edges[G[x][i]^];
if(!vis[e.from] && e.cap>e.flow)
{
vis[e.from]=;
d[e.from] = d[x]+;
Q.push(e.from);
}
}
}
return vis[s];
}
int Augment()
{
int x=t, a=INF;
while(x!=s)
{
Edge& e = edges[p[x]];
a = min(a,e.cap-e.flow);
x = edges[p[x]].from;
}
x = t;
while(x!=s)
{
edges[p[x]].flow+=a;
edges[p[x]^].flow-=a;
x=edges[p[x]].from;
}
return a;
}
int Maxflow(int _s,int _t)
{
s=_s; t=_t;
int flow = , i;
BFS();
if(d[s]>=n) return ;
memset(num,,sizeof(num));
memset(p,,sizeof(p));
FOR(i,n) if(d[i]<INF) num[d[i]]++;
int x=s;
memset(cur,,sizeof(cur));
while(d[s]<n)
{
if(x==t)
{
flow+=Augment();
x=s;
}
int ok=;
for(unsigned i=cur[x];i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(e.cap>e.flow && d[x]==d[e.to]+)
{
ok=;
p[e.to]=G[x][i];
cur[x]=i;
x=e.to;
break;
}
}
if(!ok)
{
int m=n-;
for(unsigned i=;i<G[x].size();i++)
{
Edge& e=edges[G[x][i]];
if(e.cap>e.flow) m=min(m,d[e.to]);
}
if(--num[d[x]]==) break;
num[d[x]=m+]++;
cur[x]=;
if(x!=s) x=edges[p[x]].from;
}
}
return flow;
}
}; ISAP isap; void run()
{
int n,m,u,v,c;
scanf("%d%d",&n,&m);
isap.init(n,m);
while(m--)
{
scanf("%d%d%d",&u,&v,&c);
isap.AddEdge(u,v,c);
//isap.AddEdge(v,u,c);
}
static int cas = ;
printf("Case %d: %d\n",cas++,isap.Maxflow(,n));
} int main()
{
freopen("case.txt","r",stdin);
int _;
scanf("%d",&_);
while(_--)
run();
return ;
}
 

HDU 3549 Flow Problem (最大流ISAP)的更多相关文章

  1. HDU 3549 Flow Problem (dinic模版 && isap模版)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3549 题意: 给你一个有向图,问你1到n的最大流. dinic模版 (n*n*m) #include ...

  2. hdu - 3549 Flow Problem (最大流模板题)

    http://acm.hdu.edu.cn/showproblem.php?pid=3549 Ford-Fulkerson算法. #include <iostream> #include ...

  3. hdu 3549 Flow Problem (最大流)

    裸最大流,做模板用 m条边,n个点,求最大流 #include <iostream> #include <cstdio> #include <cstring> #i ...

  4. hdu 3549 Flow Problem 最大流 Dinic

    题目链接 题意 裸的最大流. 学习参考 http://www.cnblogs.com/SYCstudio/p/7260613.html Code #include <bits/stdc++.h& ...

  5. 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/ ...

  6. 网络流 HDU 3549 Flow Problem

    网络流 HDU 3549 Flow Problem 题目:pid=3549">http://acm.hdu.edu.cn/showproblem.php?pid=3549 用增广路算法 ...

  7. 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 ...

  8. hdu 3549 Flow Problem

    题目连接 http://acm.hdu.edu.cn/showproblem.php?pid=3549 Flow Problem Description Network flow is a well- ...

  9. hdu 3549 Flow Problem Edmonds_Karp算法求解最大流

    Flow Problem 题意:N个顶点M条边,(2 <= N <= 15, 0 <= M <= 1000)问从1到N的最大流量为多少? 分析:直接使用Edmonds_Karp ...

随机推荐

  1. Unix高级环境编程—进程控制(一)

    一.函数fork #include<unistd.h> pid_t  fork(void)                                                 ...

  2. cmake的外部编译

    1 什么是外部编译 就是让源码文件和cmake生成的工程文件分开,将cmake生成的工程文件放在一个单独的目录下面. 2 怎样进行外部编译 第一,单独建立一个目录,这个目录在source code目录 ...

  3. apache vhosts 虚拟主机设置

    编辑vhosts文件:/alidata/server/httpd-2.4.10/conf/extra/httpd-vhosts.conf <VirtualHost *:80> <Lo ...

  4. StackOver上的一个wx刷新显示的例子

    import wx class MyFrame(wx.Frame): def __init__(self): wx.Frame.__init__(self, None) self.panel = wx ...

  5. 【题解】[P3557 POI2013]GRA-Tower Defense Game

    [题解][P3557 POI2013]GRA-Tower Defense Game 这道题是真的** 根据题目给的\(k\),可以知道,我们随便放塔,只要不全放一起,一定是一种合法的方案. 直接枚举就 ...

  6. 链表的C++实现

      有的时候,处于内存中的数据并非连续的.那么这时候.我们就须要在数据结构中加入一个属性.这个属性会记录以下一个数据的地址.有了这个地址之后.全部的数据就像一条链子一样串起来了,那么这个地址属性就起到 ...

  7. ABap-小技巧

    if FIELD cn '0123456789'. *&如果字符串包含‘数字’    STOP.  endif. 同理到字母‘ABCDEFG*’ 'abcdefg*' '/' '\' 等其它字 ...

  8. (转)nginx-rtmp-module和ffmpeg搭建实时HLS切片

    1.rtmp服务器 nginx+pcre+zlib+openssl+nginx-rtmp-module ./configure \ --prefix=/usr/local/nginx \ --sbin ...

  9. Linux线程的几种结束方式

    Linux创建线程使用 int pthread_create(pthread_t *thread, const pthread_attr_t *attr, void *(*start_routine) ...

  10. JS高级调试技巧:捕获和分析 JavaScript Error详解

    前端工程师都知道 JavaScript 有基本的异常处理能力.我们可以 throw new Error(),浏览器也会在我们调用 API 出错时抛出异常.但估计绝大多数前端工程师都没考虑过收集这些异常 ...