BZOJ 3894 文理分科 最小割

 对于全理/全文分别建一个点代表他 然后S->点/点->T连一条收益边 全理/全文集合里的点向它连INF的边 使得只要存在一个不是理/文 这条边就要被割掉

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef int JQK;
int n, m;
const int INF = INT_MAX;int dir[4][2] = {{0, 1}, {1, 0}, {0, -1}, { -1, 0}};
int bl[105][105];
int sum = 0;
int A[105][105], B[105][105], SA[105][105], SB[105][105];
int main() {
        int n, m;
        scanf("%d %d", &n, &m);
        dinic::MAXP = 3 * n * m + 5;
        int s, t;
        s = 3 * n * m + 1, t = s + 1;
        dinic::init(s, t);
        for (int i = 1; i <= n; i++)
                for (int j = 1; j <= m; j++) {
                        scanf("%d", &A[i][j]);
                        sum += A[i][j];
                }
        for (int i = 1; i <= n; i++)
                for (int j = 1; j <= m; j++) {
                        scanf("%d", &B[i][j]);
                        sum += B[i][j];
                }
        for (int i = 1; i <= n; i++)
                for (int j = 1; j <= m; j++) {
                        scanf("%d", &SA[i][j]);
                        sum += SA[i][j];
                }
        for (int i = 1; i <= n; i++)
                for (int j = 1; j <= m; j++) {
                        scanf("%d", &SB[i][j]);
                        sum += SB[i][j];
                }
        for (int i = 1; i <= n; i++)
                for (int j = 1; j <= m; j++) {
                        bl[i][j] = (i + j) % 2;
                }
        for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= m; j++) {
                        int now = (i - 1) * m + j;
                        dinic::addedge(s, now, A[i][j]);
                        dinic::addedge(now, t, B[i][j]);
                        dinic::addedge(s, now + n * m, SA[i][j]);
                        dinic::addedge(now + n * m, now, INF);
                        dinic::addedge(now + 2 * n * m, t, SB[i][j]);
                        dinic::addedge(now, now + 2 * n * m, INF);
                        for (int k = 0; k < 4; k++) {
                                int dx = i + dir[k][0];
                                int dy = j + dir[k][1];
                                if (dx >= 1 && dx <= n && dy >= 1 && dy <= m) {
                                        int aim = (dx - 1) * m + dy;
                                        dinic::addedge(aim, now + 2 * m * n, INF);
                                        dinic::addedge(now + n * m, aim, INF);
                                }
                        }
                }
        }
        printf("%d
", sum - dinic::Dinic());
        return 0;
}