BZOJ 2127 happiness 最小割

标签：bzoj   网络流   最小割   

题目大意：一个班级中，两个相邻的人是朋友，一对朋友如果同时选择文科或者理科会有加成。问最多能够得到的权值是多少。

思路：先假设得到所有的权值，然后运用最小割的意义求出最大的获利。具体方法见 http://hzwer.com/2422.html

CODE：

#define _CRT_SECURE_NO_WARNINGS

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 1010
#define MAXP 5000010
#define MAXE 5000010
#define INF 0x3f3f3f3f
#define S 0
#define T (MAXP - 1)
using namespace std;
#define P(i,j) ((i - 1) * (m) + (j))

struct MaxFlow{
	int head[MAXP],total;
	int next[MAXE],aim[MAXE],flow[MAXE];

	int deep[MAXP];

	MaxFlow():total(1) {}
	void Add(int x,int y,int f) {
		next[++total] = head[x];
		aim[total] = y;
		flow[total] = f;
		head[x] = total;
	}
	void Insert(int x,int y,int f) {
		Add(x,y,f);
		Add(y,x,f);
	}
	bool BFS() {
		static queue<int> q;
		while(!q.empty())	q.pop();
		memset(deep,0,sizeof(deep));
		deep[S] = 1;
		q.push(S);
		while(!q.empty()) {
			int x = q.front(); q.pop();
			for(int i = head[x]; i; i = next[i])
				if(!deep[aim[i]] && flow[i]) {
					deep[aim[i]] = deep[x] + 1;
					q.push(aim[i]);
					if(aim[i] == T)	return true;
				}
		}
		return false;
	}
	int Dinic(int x,int f) {
		if(x == T)	return f;
		int temp = f;
		for(int i = head[x]; i; i = next[i])
			if(flow[i] && temp && deep[aim[i]] == deep[x] + 1) {
				int away = Dinic(aim[i],min(flow[i],temp));
				if(!away)	deep[aim[i]] = 0;
				flow[i] -= away;
				flow[i^1] += away;
				temp -= away;
			}
		return f - temp;
	} 
}solver;

int m,n,ans;
int s[MAX][MAX],t[MAX][MAX];

int main()
{
	cin >> m >> n;
	for(int i = 1; i <= m; ++i)
		for(int x,j = 1; j <= n; ++j) {
			scanf("%d",&x);
			s[i][j] += x << 1;
			ans += (x << 1);
		}
	for(int i = 1; i <= m; ++i)
		for(int x,j = 1; j <= n; ++j) {
			scanf("%d",&x);
			t[i][j] += x << 1;
			ans += (x << 1);
		}
	for(int i = 1; i < m; ++i)
		for(int x,j = 1; j <= n; ++j) {
			scanf("%d",&x);
			solver.Insert(P(i,j),P(i + 1,j),x);
			s[i][j] += x,s[i + 1][j] += x;
			ans += x << 1;
		}
	for(int i = 1; i < m; ++i)
		for(int x,j = 1; j <= n; ++j) {
			scanf("%d",&x);
			solver.Insert(P(i,j),P(i + 1,j),x);
			t[i][j] += x,t[i + 1][j] += x;
			ans += x << 1;
		}
	for(int i = 1; i <= m; ++i)
		for(int x,j = 1; j < n; ++j) {
			scanf("%d",&x);
			solver.Insert(P(i,j),P(i,j + 1),x);
			s[i][j] += x,s[i][j + 1] += x;
			ans += x << 1;
		}
	for(int i = 1; i <= m; ++i)
		for(int x,j = 1; j < n; ++j) {
			scanf("%d",&x);
			solver.Insert(P(i,j),P(i,j + 1),x);
			t[i][j] += x,t[i][j + 1] += x;
			ans += x << 1;
		}
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j) {
			solver.Insert(S,P(i,j),s[i][j]);
			solver.Insert(P(i,j),T,t[i][j]);
		}
	int max_flow = 0;
	while(solver.BFS())
		max_flow += solver.Dinic(S,INF);
	cout << (ans - max_flow) / 2 << endl;
	return 0;
}

BZOJ 2127 happiness 最小割

标签：bzoj   网络流   最小割   

原文地址：http://blog.csdn.net/jiangyuze831/article/details/43057535