洛谷P2774 方格取数问题(最小割)

题意

$n \times m$的矩阵，不能取相邻的元素，问最大能取多少

Sol

首先补集转化一下：最大权值 = sum - 使图不连通的最小权值

进行黑白染色

从S向黑点连权值为点权的边

从白点向T连权值为点券的边

黑点向白点连权值为INF的边

这样就转化成了最小割问题，跑Dinic即可

/*
首先补集转化一下：最大权值 = sum - 使图不连通的最小权值
进行黑白染色
从S向黑点连权值为点权的边
从白点向T连权值为点券的边
黑点向白点连权值为INF的边
跑Dinic
*/
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 10;
inline int read() {
    char c = getchar();
    int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = 1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, M, a[101][101], black[101][101], S, T = 100001;
struct Edge {
    int u, v, f, nxt;
} E[MAXN];
int head[MAXN], cur[MAXN], num = 0;
inline void add_edge(int x, int y, int z) {
    E[num] = (Edge) {x, y, z, head[x]};
    head[x] = num++;    
}
inline void AddEdge(int x, int y, int z) {
    add_edge(x, y, z);
    add_edge(y, x, 0);
}
int trans(int x, int y) {
    return (x - 1) * M + y;
}
int xx[5] = {0, -1, +1, 0, 0};
int yy[5] = {0, 0, 0, -1, +1};
int deep[MAXN];
inline int BFS() {
    queue<int> q; q.push(S);
    memset(deep, 0, sizeof(deep));
    deep[S] = 1;
    while(!q.empty()) {
        int p = q.front(); q.pop();
        for(int i = head[p]; i != -1; i = E[i].nxt) {
            int to = E[i].v;
            if(!deep[to] && E[i].f) 
                deep[to] = deep[p] + 1, q.push(to);
        }
    }
    return deep[T];
}
int DFS(int x, int flow) {
    if(x == T) return flow;
    int ansflow = 0;
    for(int &i = cur[x]; i != -1; i = E[i].nxt) {
        int to = E[i].v;
        if(E[i].f && deep[to] == deep[x] + 1) {
            int nowflow = DFS(to, min(flow, E[i].f));
            E[i].f -= nowflow; E[i ^ 1].f += nowflow;
            flow -= nowflow; ansflow += nowflow;
            if(flow <= 0) break;
        }
    }
    return ansflow;
}
int Dinic() {
    int ans = 0;
    while(BFS()) {
        memcpy(cur, head, sizeof(head));
        ans += DFS(S, INF);
    }
    return ans;
}
int main() {
    memset(head, -1, sizeof(head));
    N = read();
    M = read();
    int sum = 0;
    for(int i = 1; i <= N; i++)
        for(int j = 1; j <= M; j++) {
            a[i][j] = read(); sum += a[i][j];
            if((i + j) % 2 == 0) {
                AddEdge(S, trans(i, j), a[i][j]);    
                for(int k = 1; k <= 4; k++) {
                    int wx = i + xx[k], wy = j + yy[k];
                    if(wx >= 1 && wx <= N && wy >= 1 && wy <= M)
                        AddEdge(trans(i, j), trans(wx, wy), INF);
                }
            }
            else AddEdge(trans(i, j), T, a[i][j]);
        }
    printf("%d", sum - Dinic());
    return 0;
}
/*
3 3
1 000 00-
1 00- 0-+
2 0-- -++

*/