[SWUST1744] 方格取数问题（最大流，最大独立集）

题目链接：https://www.oj.swust.edu.cn/problem/show/1744

希望取到的点都是不相邻的（相邻：四连通），那么可以用二分图表示相连关系，然后求最大独立集。

最大独立集就是取的点均不相连，并且权值最大。

给两部分的点从1到n*n标号，源汇点分别连点的容量是对应点的权值，相邻点之间的容量为inf。

用点总权值-最小割即为最大独立集。

#include <bits/stdc++.h>
using namespace std;

typedef struct Edge {
    int u, v, w, next;
}Edge;

const int inf = 0x7f7f7f7f;
const int maxn = 9090;

int cnt, dhead[maxn];
int cur[maxn], dd[maxn];
Edge dedge[maxn<<3];
int S, T, N;

void init() {
    memset(dhead, -1, sizeof(dhead));
    for(int i = 0; i < maxn; i++) dedge[i].next = -1;
    S = 0; cnt = 0;
}

void adde(int u, int v, int w, int c1=0) {
    dedge[cnt].u = u; dedge[cnt].v = v; dedge[cnt].w = w; 
    dedge[cnt].next = dhead[u]; dhead[u] = cnt++;
    dedge[cnt].u = v; dedge[cnt].v = u; dedge[cnt].w = c1; 
    dedge[cnt].next = dhead[v]; dhead[v] = cnt++;
}

bool bfs(int s, int t, int n) {
    queue<int> q;
    for(int i = 0; i < n; i++) dd[i] = inf;
    dd[s] = 0;
    q.push(s);
    while(!q.empty()) {
        int u = q.front(); q.pop();
        for(int i = dhead[u]; ~i; i = dedge[i].next) {
            if(dd[dedge[i].v] > dd[u] + 1 && dedge[i].w > 0) {
                dd[dedge[i].v] = dd[u] + 1;
                if(dedge[i].v == t) return 1;
                q.push(dedge[i].v);
            }
        }
    }
    return 0;
}

int dinic(int s, int t, int n) {
    int st[maxn], top;
    int u;
    int flow = 0;
    while(bfs(s, t, n)) {
        for(int i = 0; i < n; i++) cur[i] = dhead[i];
        u = s; top = 0;
        while(cur[s] != -1) {
            if(u == t) {
                int tp = inf;
                for(int i = top - 1; i >= 0; i--) {
                    tp = min(tp, dedge[st[i]].w);
                }
                flow += tp;
                for(int i = top - 1; i >= 0; i--) {
                    dedge[st[i]].w -= tp;
                    dedge[st[i] ^ 1].w += tp;
                    if(dedge[st[i]].w == 0) top = i;
                }
                u = dedge[st[top]].u;
            }
            else if(cur[u] != -1 && dedge[cur[u]].w > 0 && dd[u] + 1 == dd[dedge[cur[u]].v]) {
                st[top++] = cur[u];
                u = dedge[cur[u]].v;
            }
            else {
                while(u != s && cur[u] == -1) {
                    u = dedge[st[--top]].u;
                }
                cur[u] = dedge[cur[u]].next;
            }
        }
    }
    return flow;
}

typedef pair<int, int> pii;
const int dx[5] = {0, 0, 1, -1};
const int dy[5] = {1, -1, 0, 0};
int n, m, icnt;
int G[33][33];
map<pii, int> wobuzhidao;

inline int id(int x, int y) {
    return wobuzhidao[pii(x, y)];
}

inline bool ok(int x, int y) {
    return x >= 1 && x <= n && y >= 1 && y <= m;
}

int main() {
    // freopen("in", "r", stdin);
    int u, v, w;
    while(~scanf("%d%d",&n,&m)) {
        init(); wobuzhidao.clear(); icnt = 0;
        memset(G, 0, sizeof(G));
        int sum = 0;
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                wobuzhidao[pii(i, j)] = ++icnt;
                scanf("%d", &G[i][j]);
                sum += G[i][j];
            }
        }
        S = 0, T = icnt * 2 + 1, N = T + 1;
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                adde(S, id(i,j), G[i][j]);
                adde(icnt+id(i,j), T, G[i][j]);
                for(int k = 0; k < 4; k++) {
                    int x = i + dx[k];
                    int y = j + dy[k];
                    if(!ok(x, y)) continue;
                    adde(id(i,j), icnt+id(x,y), inf);
                }
            }
        }
        int ret = dinic(S, T, N);
        printf("%d
", sum - ret / 2);
    }
    return 0;
}