BZOJ1475 方格取数 <黑白染色+最小割>

Problem

方格取数

Time Limit: $5 Sec$
Memory Limit: $64 MB$

Description

在一个$n\times n$的方格里，每个格子里都有一个正整数。从中取出若干数，使得任意两个取出的数所在格子没有公共边，且取出的数的总和尽量大。

Input

第一行一个数$n$（$n\le 30$），接下来$n$行每行$n$个数描述一个方阵

Output

仅一个数，即最大和

Sample Input

2
1 2
3 5

Sample Output

6

标签：黑白染色 最小割

Solution

将网格图黑白染色，同色格子间不会冲突。
建模：$S\to 黑格\ \ 容量val$，$白格\to T\ \ 容量val$，$黑格\to 白格\ \ 容量\infty$
$ans = 总权值-最小割$

Code

#include <bits/stdc++.h>
#define MAX_N 1000
#define INF 0x7f7f7f7f
using namespace std;
int nxt[4][2] = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}}, sum;
int n, m, s, t, cnt, id[35][35], a[MAX_N+5], d[MAX_N+5], pr[MAX_N+5];
struct node {int u, v, c, nxt;} E[MAX_N*MAX_N*2];
void init() {s = 0, t = m+1; memset(pr, -1, sizeof pr);}
void getID() {for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) id[i][j] = (i-1)*n+j;}
void insert(int u, int v, int c) {E[cnt].v = v, E[cnt].c = c, E[cnt].nxt = pr[u], pr[u] = cnt++;}
void addedge(int u, int v, int c) {insert(u, v, c), insert(v, u, 0);}
bool BFS() {
    queue <int> que;	que.push(s);
    memset(d, -1, sizeof d), d[s] = 0;
    while (!que.empty()) {
        int u = que.front();	que.pop();
        for (int i = pr[u]; ~i; i = E[i].nxt) {
            int v = E[i].v, c = E[i].c;
            if (!c || ~d[v])	continue;
            d[v] = d[u]+1, que.push(v);
        }
    }
    return ~d[t];
}
int DFS(int u, int flow) {
    if (u == t)	return flow;
    int ret = 0;
    for (int i = pr[u]; ~i; i = E[i].nxt) {
        int v = E[i].v, c = E[i].c;
        if (!c || d[v] != d[u]+1)	continue;
        int tmp = DFS(v, min(flow, c));
        E[i].c -= tmp, E[i^1].c += tmp;
        flow -= tmp, ret += tmp;
        if (!flow)	break;
    }
    if (!ret) d[u] = -1;
    return ret;
}
int Dinic() {int ret = 0;    while (BFS()) ret += DFS(s, INF);	return ret;}
int main() {
    scanf("%d", &n), m = n*n, getID(), init();
    for (int i = 1; i <= m; i++) scanf("%d", a+i), sum += a[i];
    for (int i = 1; i <= m; i++)
        if ((i-1)%n%2 == (i-1)/n%2) addedge(s, i, a[i]);
        else addedge(i, t, a[i]);
    for (int i = 1; i <= m; i++) if ((i-1)%n%2 == (i-1)/n%2){
        int x = (i-1)/n+1, y = (i-1)%n+1;
        for (int j = 0; j < 4; j++) {
            int nx = x+nxt[j][0], ny = y+nxt[j][1];
            if (nx < 1 || nx > n || ny < 1 || ny > n) continue;
            addedge(id[x][y], id[nx][ny], INF);
        }
    }
    printf("%d", sum-Dinic());
    return 0;
}