BZOJ 1324 Exca 王者之剑 最小割

题目大意：给出一个带权值的矩阵，取走一个地方的权值之后，与其相邻的格子的权值就会变成0，问最多可以取出多少权值。

思路：Amber论文里的题。建图不难，把图染色，然后一种颜色从S连边，另一种颜色向T连边。再把相邻的格子连边，之后跑最小割，用总权值减去最大流就是答案。

CODE：

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAX 400
#define MAXP 40000
#define MAXE 500010
#define INF 0x3f3f3f3f
#define S 0
#define T 39999
using namespace std;
const int dx[] = {0,1,0,-1,0};
const int dy[] = {0,0,1,0,-1};

int m,n,cnt;
int src[MAX][MAX],num[MAX][MAX];

int head[MAXP],total = 1;
int next[MAXE],aim[MAXE],flow[MAXE];

int deep[MAXP];

inline void Add(int x,int y,int f)
{
	next[++total] = head[x];
	aim[total] = y;
	flow[total] = f;
	head[x] = total;
}

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(flow[i] && !deep[aim[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] && deep[aim[i]] == deep[x] + 1 && temp) {
			int away = Dinic(aim[i],min(temp,flow[i]));
			if(!away)	deep[aim[i]] = 0;
			flow[i] -= away;
			flow[i^1] += away;
			temp -= away;
		}
	return f - temp;
}

int main()
{
	cin >> m >> n;
	int sum = 0;
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j) {
			scanf("%d",&src[i][j]);
			sum += src[i][j];
			num[i][j] = ++cnt;
			if((i + j)&1)
				Add(S,num[i][j],src[i][j]),Add(num[i][j],S,0);
			else
				Add(num[i][j],T,src[i][j]),Add(T,num[i][j],0);
		}
	for(int i = 1; i <= m; ++i)
		for(int j = 1; j <= n; ++j) {
			if(((i + j)&1) == 0)	continue;
			for(int k = 1; k <= 4; ++k) {
				int fx = i + dx[k];
				int fy = j + dy[k];
				if(!fx || !fy || fx > m || fy > n)	continue;
				Add(num[i][j],num[fx][fy],INF);
				Add(num[fx][fy],num[i][j],0);
			}
		}
	int max_flow = 0;
	while(BFS())
		max_flow += Dinic(S,INF);
	cout << sum - max_flow << endl;
	return 0;
}