First problem to learn Max Flow.

Ford-Fulkerson is a group of algorithms - Dinic is one of it.
It is an iterative process: we use BFS to check augament-ability, and use DFS to augment it.

Here is the code with my comments from its tutorial

#include <cmath>
#include <climits>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <iostream>
#include <algorithm>
using namespace std; /*
* Graph Model
*/
const int MAXA = ;
const int MAXV = ; int A, V, source, dest;
// index based logging
int cap[MAXA], flow[MAXA], ady[MAXA], nexts[MAXA], last[MAXV];
int now[MAXA], level[MAXV]; void ADD(int u, int v, int c) // from u to v with cap c
{
// + edge
cap[A] = c; flow[A] = ;
ady[A] = v; nexts[A] = last[u]; last[u] = A++;
// - edge
cap[A] = ; flow[A] = ;
ady[A] = u; nexts[A] = last[v]; last[v] = A++;
} /*
* Dinic Algorithm
*/
bool BFS(int source, int dest)
{
memset(level, -, sizeof(level));
level[source] = ; queue<int> q;
q.push(source);
while (!q.empty() && level[dest] == -)
{
int u = q.front(); q.pop(); // from
for (int i = last[u]; i != -; i = nexts[i])
{
int v = ady[i]; // to
if (level[v] == - && flow[i] < cap[i])
{
level[v] = level[u] + ; // mark level
q.push(v);
}
} }
return level[dest] != -;
} int DFS(int u, int aux)
{
if (u == dest) return aux; for (int i = now[u]; i != -; now[u] = i = nexts[i])
{
int v = ady[i];
// next aux-able level node
if (level[v] > level[u] && flow[i] < cap[i])
{
int ret = DFS(v, min(aux, cap[i] - flow[i]));
if (ret > )
{
flow[i] += ret; // + edge
flow[i ^ ] -= ret;// - edge
return ret;
}
}
}
return ;
} long long Dinic()
{
long long flow = , aum;
while (BFS(source, dest))
{
for (int i = ; i <= V; i++) now[i] = last[i];
while ((aum = DFS(source, INT_MAX)) > ) flow += aum;
}
return flow;
} /*
*
*/
int main()
{
// [index]: first n is cluster, next m is wizard..
memset(last, -, sizeof(last)); int n, m, v, cc;
cin >> n >> m; source = ;
V = dest = n + m + ; // 1. Source -> Cluster with No. with people
// Cluster -> Dest with cap of 1 - means no transform
// no. of people of each skill
for (int i = ; i <= n; i++)
{
cin >> v;
if (v) ADD(source, i, v);
ADD(i, dest, ); // a non-transformed edge
}
// wizard info
for (int i = ; i <= m; i++) // m wizards
{
// array A - index of from-skill
cin >> cc;
for (int j = ; j < cc; j++)
{
cin >> v;
ADD(v, n + i, ); // skill[v](from) -> wizard[i]
}
// array B - index of to-skill
cin >> cc;
for (int j = ; j < cc; j++)
{
cin >> v;
ADD(n + i, v, ); // wizard[i] -> skill[v](to)
}
} cout << Dinic() << endl;
return ;
}

HackerRank "Training the army" - Max Flow的更多相关文章

  1. BZOJ 4390: [Usaco2015 dec]Max Flow

    4390: [Usaco2015 dec]Max Flow Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 177  Solved: 113[Submi ...

  2. 洛谷P3128 [USACO15DEC]最大流Max Flow [树链剖分]

    题目描述 Farmer John has installed a new system of  pipes to transport milk between the  stalls in his b ...

  3. Max Flow

    Max Flow 题目描述 Farmer John has installed a new system of N−1 pipes to transport milk between the N st ...

  4. min cost max flow算法示例

    问题描述 给定g个group,n个id,n<=g.我们将为每个group分配一个id(各个group的id不同).但是每个group分配id需要付出不同的代价cost,需要求解最优的id分配方案 ...

  5. [Luogu 3128] USACO15DEC Max Flow

    [Luogu 3128] USACO15DEC Max Flow 最近跟 LCA 干上了- 树剖好啊,我再也不想写倍增了. 以及似乎成功转成了空格选手 qwq. 对于每两个点 S and T,求一下 ...

  6. [Usaco2015 dec]Max Flow 树上差分

    [Usaco2015 dec]Max Flow Time Limit: 10 Sec  Memory Limit: 128 MBSubmit: 353  Solved: 236[Submit][Sta ...

  7. 洛谷P3128 [USACO15DEC]最大流Max Flow

    P3128 [USACO15DEC]最大流Max Flow 题目描述 Farmer John has installed a new system of N-1N−1 pipes to transpo ...

  8. BZOJ4390: [Usaco2015 dec]Max Flow

    BZOJ4390: [Usaco2015 dec]Max Flow Description Farmer John has installed a new system of N−1 pipes to ...

  9. P3128 [USACO15DEC]最大流Max Flow(LCA+树上差分)

    P3128 [USACO15DEC]最大流Max Flow 题目描述 Farmer John has installed a new system of  pipes to transport mil ...

随机推荐

  1. Linux 常用命令笔记

    Linux 常用命令笔记 1. locate locate:用来定位文件的位置,如:locate a.txt 但是这个命令有延迟,也就是新建的文件不一定能搜索到,如果非要找到新建的文件可以使用 upd ...

  2. PAT (Basic Level) Practise:1040. 有几个PAT

    [题目链接] 字符串APPAPT中包含了两个单词“PAT”,其中第一个PAT是第2位(P),第4位(A),第6位(T):第二个PAT是第3位(P),第4位(A),第6位(T). 现给定字符串,问一共可 ...

  3. java 读取文件的常用方式

    1.读取: public class ReadFromFile { /** * 以字节为单位读取文件,常用于读二进制文件,如图片.声音.影像等文件. */ public static void rea ...

  4. bedtools 每天都会用到的工具

    详细的使用说明:http://bedtools.readthedocs.org/en/latest/ Collectively, the bedtools utilities are a swiss- ...

  5. MySql配置方法,批处理进行MySql配置

    @echo 注意右击管理员运行 @pause @echo .添加环境变量 @echo .my.ini中添加如下配置 @echo [mysqld] @echo basedir=C:\Program Fi ...

  6. man page的介绍

    假设你知道一个命令名,但不知其使用方法,此时可以求助于在线求助系统,即使用man(manual)命令,如man date. 上图为执行man date语句后的结果.<翻页查看时可按空格键,退出m ...

  7. HOG特征(Histogram of Gradient)总结(转载)

    整理一下我个人觉得比较好的HOG博文 博文1:OpenCV HOGDescriptor: 参数与图解 http://blog.csdn.NET/raodotcong/article/details/6 ...

  8. 162. Find Peak Element

    A peak element is an element that is greater than its neighbors. Given an input array where num[i] ≠ ...

  9. 批量kill相关所有进程

    首先,用ps查看进程,方法如下: $ ps -ef …… smx       1822     1  0 11:38 ?        00:00:49 gnome-terminal smx      ...

  10. POJ 2186 Popular Cows(强连通)

                                                                  Popular Cows Time Limit: 2000MS   Memo ...