Flow Problem

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
 

————————————————————————————————————————————————————————


题目要求很简单,求从第一个点到最后一个点的最大流量。
最大流板子题,直接套模板,

Dinic
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset> using namespace std; #define LL long long
const int INF = 0x3f3f3f3f;
#define MAXN 500 struct node
{
int u, v, next, cap;
} edge[MAXN*MAXN];
int nt[MAXN], s[MAXN], d[MAXN], visit[MAXN];
int cnt; void init()
{
cnt = 0;
memset(s, -1, sizeof(s));
} void add(int u, int v, int c)
{
edge[cnt].u = u;
edge[cnt].v = v;
edge[cnt].cap = c;
edge[cnt].next = s[u];
s[u] = cnt++;
edge[cnt].u = v;
edge[cnt].v = u;
edge[cnt].cap = 0;
edge[cnt].next = s[v];
s[v] = cnt++;
} bool BFS(int ss, int ee)
{
memset(d, 0, sizeof d);
d[ss] = 1;
queue<int>q;
q.push(ss);
while (!q.empty())
{
int pre = q.front();
q.pop();
for (int i = s[pre]; ~i; i = edge[i].next)
{
int v = edge[i].v;
if (edge[i].cap > 0 && !d[v])
{
d[v] = d[pre] + 1;
q.push(v);
}
}
}
return d[ee];
} int DFS(int x, int exp, int ee)
{
if (x == ee||!exp) return exp;
int temp,flow=0;
for (int i = nt[x]; ~i ; i = edge[i].next, nt[x] = i)
{
int v = edge[i].v;
if (d[v] == d[x] + 1&&(temp = (DFS(v, min(exp, edge[i].cap), ee))) > 0)
{
edge[i].cap -= temp;
edge[i ^ 1].cap += temp;
flow += temp;
exp -= temp;
if (!exp) break;
}
}
if (!flow) d[x] = 0;
return flow;
} int Dinic_flow(int ss, int ee)
{
int ans = 0;
while (BFS(ss, ee))
{
for (int i = 0; i <= ee; i++) nt[i] = s[i];
ans+= DFS(ss, INF, ee);
}
return ans;
} int main()
{
int n,m,u,v,c,T;
int q=1;
scanf("%d",&T);
while (T--)
{
scanf("%d %d", &n, &m);
init();
for(int i=0; i<m; i++)
{
scanf("%d%d%d",&u,&v,&c);
add(u,v,c);
}
printf("Case %d: %d\n",q++,Dinic_flow(1,n)); }
return 0;
}

isap+bfs
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <queue>
#include <vector>
#include <stack>
#include <set>
#include <map>
using namespace std;
const int MAXN =100010;//点max
const int MAXM=400010;//边max
const int INF=0x3f3f3f3f;
struct Edge
{
int to,next,cap,flow;
} edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],cur[MAXN];
void init()
{
tol=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw=0)
{
edge[tol].to=v;
edge[tol].cap=w;
edge[tol].flow=0;
edge[tol].next=head[u];
head[u]=tol++;
edge[tol].to=u;
edge[tol].cap=rw;
edge[tol].flow=0;
edge[tol].next=head[v];
head[v]=tol++;
} int Q[MAXN];
void BFS(int start,int end)
{
memset(dep,-1,sizeof(dep));
memset(gap,0,sizeof(gap));
gap[0]=1;
int front=0,rear=0;
dep[end]=0;
Q[rear++]=end;
while(front!=rear)
{
int u=Q[front++];
for(int i=head[u]; i!=-1; i=edge[i].next)
{
int v=edge[i].to;
if(dep[v]!=-1)
continue;
Q[rear++]=v;
dep[v]=dep[u]+1;
gap[dep[v]]++;
}
}
} int S[MAXN];
int sap(int start,int end,int N)
{
BFS(start,end);
memcpy(cur,head,sizeof(head));
int top=0;
int u =start;
int ans=0;
while(dep[start]<N)
{
if(u==end)
{
int Min=INF;
int inser;
for(int i=0; i<top; i++)
{
if(Min>edge[S[i]].cap-edge[S[i]].flow)
{
Min=edge[S[i]].cap-edge[S[i]].flow;
inser=i;
}
} for(int i=0; i<top; i++)
{
edge[S[i]].flow+=Min;
edge[S[i]^1].flow-=Min; }
ans+=Min;
top=inser;
u=edge[S[top]^1].to;
continue;
} bool flag=false;
int v;
for(int i=cur[u]; i!=-1; i=edge[i].next)
{
v=edge[i].to;
if(edge[i].cap-edge[i].flow&&dep[v]+1==dep[u])
{
flag=true;
cur[u]=i;
break;
} }
if(flag)
{
S[top++]=cur[u];
u=v;
continue;
}
int Min=N;
for(int i=head[u]; i!=-1; i=edge[i].next)
{
if(edge[i].cap-edge[i].flow&&dep[edge[i].to]<Min)
{
Min=dep[edge[i].to];
cur[u]=i;
}
}
gap[dep[u]]--;
if(!gap[dep[u]])return ans;
dep[u]=Min+1;
gap[dep[u]]++;
if(u!=start)
u=edge[S[--top]^1].to;
}
return ans; } int main()
{
int n,m,u,v,w,T;
while(~scanf("%d",&T))
{
int q=0;
while(T--)
{
scanf("%d%d",&m,&n);
init();
for(int i=0; i<n; i++)
{
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w,0);
}
int ans=sap(1,m,m);
printf("Case %d: %d\n",++q,ans);
}
} return 0;
}

Hdu3549 Flow Problem 2017-02-11 16:24 58人阅读 评论(0) 收藏的更多相关文章

  1. Hardwood Species 分类: POJ 树 2015-08-05 16:24 2人阅读 评论(0) 收藏

    Hardwood Species Time Limit: 10000MS Memory Limit: 65536K Total Submissions: 20619 Accepted: 8083 De ...

  2. hdu 1503, LCS variants, find a LCS, not just the length, backtrack to find LCS, no extra markup 分类: hdoj 2015-07-18 16:24 139人阅读 评论(0) 收藏

    a typical variant of LCS algo. the key point here is, the dp[][] array contains enough message to de ...

  3. 团体程序设计天梯赛L3-010 是否完全二叉搜索树 2017-03-24 16:12 29人阅读 评论(0) 收藏

    L3-010. 是否完全二叉搜索树 时间限制 400 ms 内存限制 65536 kB 代码长度限制 8000 B 判题程序 Standard 作者 陈越 将一系列给定数字顺序插入一个初始为空的二叉搜 ...

  4. Mahout快速入门教程 分类: B10_计算机基础 2015-03-07 16:20 508人阅读 评论(0) 收藏

    Mahout 是一个很强大的数据挖掘工具,是一个分布式机器学习算法的集合,包括:被称为Taste的分布式协同过滤的实现.分类.聚类等.Mahout最大的优点就是基于hadoop实现,把很多以前运行于单 ...

  5. hdu 1082, stack emulation, and how to remove redundancy 分类: hdoj 2015-07-16 02:24 86人阅读 评论(0) 收藏

    use fgets, and remove the potential '\n' in the string's last postion. (main point) remove redundanc ...

  6. Least Common Ancestors 分类: ACM TYPE 2014-10-19 11:24 84人阅读 评论(0) 收藏

    #include <iostream> #include <cstdio> #include <cstring> #include <cmath> #i ...

  7. C语言之void类型及void指针 分类: C/C++ 2015-07-13 11:24 8人阅读 评论(0) 收藏

    原文网址:http://www.cnblogs.com/pengyingh/articles/2407267.html 1.概述 许多初学者对C/C 语言中的void及void指针类型不甚理解,因此在 ...

  8. 初学Larevel 2014-08-21 11:24 90人阅读 评论(0) 收藏

    添加第一个路由时就遇到了 404错误,查了一下说要这样才能 版权声明:本文为博主原创文章,未经博主允许不得转载.

  9. 网上关于sort结构体排序都不完整,我来写一个完整版的 2014-08-09 16:50 60人阅读 评论(0) 收藏

    主要参考sort函数_百度文库, 但是那篇有错误 2.结构体排序,a升,b降,c降 平板视图 打印? 01 #include <iostream> 02 #include <algo ...

随机推荐

  1. WCF 快速入门

    定义服务契约 构建HelloWCF应用的第一步是创建服务契约.契约式是表示消息应用外形的主要方式.对于外形,是指服务暴露的操作,使用的消息 schema和每个操作实现的消息交换模式(MEP).总之,契 ...

  2. EF-CodeFirst系列100

    .学习资料地址: CodeFirst:https://msdn.microsoft.com/zh-cn/data/jj193542 一.CodeFirst迁移(Migration) .工具--> ...

  3. configure: error: Please reinstall the libcurl distribution - easy.h should be in <curl-dir>/include/curl/

    编译php出现错误: configure: error: Please reinstall the libcurl distribution - easy.h should be in <cur ...

  4. delphi 触摸 手势

    delphi手势,左右滑动, 控件的OnGesture事件写代码. 放一个TGestureManager控件,设置控件的touch属性为TGestureManager控件. 然后勾选控件的Touch& ...

  5. javascript常用验证大全

    1. 长度限制 <script> function test() { if(document.a.b.value.length>50) { alert("不能超过50个字符 ...

  6. google-gson库下的gson的基本使用

    public class Users { private String username; private String password; private Integer age; public S ...

  7. WSAEvent

    [WSAEvent] 1.WSACreateEvent.创建WSA事件. 2.WSAEventSelect. lNetworkEvents [in] A bitmask that specifies ...

  8. 【转】字符串匹配的KMP算法:移动位数 = 已匹配 - 部分匹配值(共有长度)

    计算机的基本任务之一. 举例来说,有一个字符串"BBC ABCDAB ABCDABCDABDE",我想知道,里面是否包含另一个字符串"ABCDABD"? 许多算 ...

  9. 修改数据库的instance_name和db_name

    分成两个步骤,先修改instance_name,在修改db_name 修改SID1.全备份数据库RMAN> backup as compressed backupset database inc ...

  10. linux环境下pdo加载问题

    报错信息信息 PHP Warning: PHP Startup: Unable to load dynamic library ‘/usr/lib/php5/20121212/pdo_mysql.so ...