BZOJ 2127 happiness 最小割

来源：互联网发布：表单提交json数据编辑：程序博客网时间：2024/05/22 18:56

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;}

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