Dinic

hdoj 1569 & hdoj 1565 方格取数

1. 最小点权覆盖集=最小割=最大流
2. 最大点权独立集=总权-最小点权覆盖集

将格子黑白染色，然后求最小割。

#include <iostream>
#include <cstring>
using namespace std;

/*==================================================*\
| Dinic 最大流 O(V^2 * E)
| INIT: ne=2; head[]置为0; addedge()加入所有弧;
| CALL: flow(n, s, t);
\*==================================================*/
//吉大模版
#define typec int // type of cost
#define E 105005
#define N 2505
//const int N=2500;
//const int E=300000;
const typec inf = 0x3f3f3f3f; // max of cost
struct edge { int x, y, nxt; typec c; } bf[E];
int ne, head[N], cur[N], ps[N], dep[N];


void addedge(int x, int y, typec c)
{ // add an arc(x -> y, c); vertex: 0 ~ n-1;
	bf[ne].x = x; bf[ne].y = y; bf[ne].c = c;
	bf[ne].nxt = head[x]; head[x] = ne++;
	bf[ne].x = y; bf[ne].y = x; bf[ne].c = 0;
	bf[ne].nxt = head[y]; head[y] = ne++;
}


typec flow(int n, int s, int t)
{
	typec tr, res = 0;
	int i, j, k, f, r, top;
	while (1) {
		memset(dep, -1, n * sizeof(int));
		for (f = dep[ps[0] = s] = 0, r = 1; f != r; )
			for (i = ps[f++], j = head[i]; j; j = bf[j].nxt)
			{
				if (bf[j].c && -1 == dep[k = bf[j].y]){
					dep[k] = dep[i] + 1; ps[r++] = k;
					if (k == t) { f = r; break; }
				}
			}
			if (-1 == dep[t]) break;
			memcpy(cur, head, n * sizeof(int));
			for (i = s, top = 0; ; ) {
				if (i == t) {
					for (k = 0, tr = inf; k < top; ++k)
						if (bf[ps[k]].c < tr)
							tr = bf[ps[f = k]].c;
						for (k = 0; k < top; ++k)
							bf[ps[k]].c -= tr, bf[ps[k]^1].c += tr;
						res += tr; i = bf[ps[top = f]].x;
				}
				for (j=cur[i]; cur[i]; j = cur[i] = bf[cur[i]].nxt)
					if (bf[j].c && dep[i]+1 == dep[bf[j].y]) break;
					if (cur[i]) {
						ps[top++] = cur[i];
						i = bf[cur[i]].y;
					}
					else {
						if (0 == top) break;
						dep[i] = -1; i = bf[ps[--top]].x;
					}
			}
	}
	return res;
}



int main() {
//	freopen( "c:/aaa.txt", "r", stdin );
	int i, j, k, a, s, t, n, m, x, y, sum;
	int dir[4][2] = { 0, 1, 1, 0 , 0, -1, -1, 0 };

	while( scanf( "%d", &n) == 1 ) {
		m = n;
		ne = 2;
		memset( head, 0, sizeof(head) );
		s = n * m;
		t = s + 1;
		sum = 0;
		for( i=0; i<n; ++i ) {
			for( j=0; j<m; ++j ) {
				scanf( "%d", &a );
				sum += a;
				if( (i+j)%2 == 0 ) {
					addedge( s, i*m+j, a );
					for( k=0; k<4; ++k ) {
						x = i + dir[k][0];
						y = j + dir[k][1];
						if( x < 0 || y < 0 || x >= n || y >= m ) continue;
						addedge( i*m+j, x*m+y, inf );
					}
				} else {
					addedge( i*m+j, t, a );
				}
			}
		}
		int rec = flow(n*m+2, s, t);

		printf( "%d\n", sum - rec);
	}
	return 0;
}

				  

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