Luogu P4313 文理分科

题目传送门

好久没做网络流了……

---

定义四个矩阵分别为a[i][j], b[i][j], c[i][j], d[i][j]
定义一个点为(x)，源点向(x)连容量为a[i][j]边，(x)向汇点连容量为b[i][j]的边
因为相邻的选择同一科目会有加成，所以对于c[i][j]，我们可以新建一个节点(y)，源点向(y)连容量为c[i][j]的边，(y)向它周围的点连容量为(inf)的边。对于d[i][j]也是同理，(y)向汇点连容量为c[i][j]的边，周围的点向(y)连容量为(inf)的边
将四个矩阵的价值全加起来，减去最小割即为答案

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
#define LL long long
#define INF 2147483647
int read() {
    int k = 0, f = 1; char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        k = k * 10 + c - 48, c = getchar();
    }
    return k * f;
}
struct zzz {
    int t, len, nex;
}e[500010 << 1]; int head[100010], tot = 1;
void add(int x, int y, int z) {
    e[++tot].t = y;
    e[tot].len = z;
    e[tot].nex = head[x];
    head[x] = tot;

    e[++tot].t = x;
    e[tot].len = 0;
    e[tot].nex = head[y];
    head[y] = tot;
}
int s, t, n, m, vis[100010];
bool bfs() {
    queue <int> q; q.push(s);
    memset(vis, 0, sizeof(vis)); vis[s] = 1;
    while(!q.empty()) {
        int k = q.front(); q.pop();
        for(int i = head[k]; i; i = e[i].nex) {
            if(!vis[e[i].t] && e[i].len) {
                vis[e[i].t] = vis[k] + 1; q.push(e[i].t);
                if(e[i].t == t) return 1;
            }
        }
    }
    return vis[t];
}
int dfs(int flow, int pos) {
    if(!flow || pos == t) return flow;
    int fl, rest = 0;
    for(int i = head[pos]; i; i = e[i].nex) {
        if(vis[e[i].t] == vis[pos] + 1 && (fl = dfs(min(e[i].len, flow-rest), e[i].t))) {
            e[i].len -= fl, e[i^1].len += fl, rest += fl;
            if(rest == flow) return rest;
        }
    }
    if(rest < flow) vis[pos] = 0;
    return rest;

}
int dinic() {
    int ans = 0;
    while(bfs()) ans += dfs(INF, s);
    return ans;
}
inline int calc(int x, int y) {
    return (x - 1) * m + y + 2;
}
int fx[5] = {0, 1, 0, -1, 0},
    fy[5] = {0, 0, 1, 0, -1};
int sum;
int main() {
    n = read(), m = read(), s = 1, t = 2;
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) {
            int x = read(); sum += x;
            add(s, calc(i, j), x);
        }
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) {
            int x = read(); sum += x;
            add(calc(i, j), t, x);
        }
    int pos = calc(n, m);
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) {
            int val = read();
            sum += val;
            add(s, ++pos, val);
            for(int k = 0; k <= 4; ++k) {
                if(i+fx[k] < 1 || i+fx[k] > n || j+fy[k] < 1 || j+fy[k] > m) continue;
                int y = calc(i+fx[k], j+fy[k]);
                add(pos, y, INF);
            }
        }
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= m; ++j) {
            int val = read();
            sum += val;
            add(++pos, t, val);
            for(int k = 0; k <= 4; ++k) {
                if(i+fx[k] < 1 || i+fx[k] > n || j+fy[k] < 1 || j+fy[k] > m) continue;
                int y = calc(i+fx[k], j+fy[k]);
                add(y, pos, INF);
            }
        }
    printf("%d", sum - dinic());
    return 0;
}