[BZOJ4808] 马（最大独立集，最大流）

题目链接：http://www.lydsy.com/JudgeOnline/problem.php?id=4808

题意：其实就是找出一个点集的子集，使得这个子集中的点互不相连。求这个子集规模最大。

就是最大独立集。点好多，有200*200个。所以用dinic优化了下。

最大独立集=N-最大匹配，最大匹配=最大流，所以最大独立集=N-最大流。

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

//下标从0开始
typedef struct Edge {
    int u, v, w, next;
}Edge;

const int inf = 0x7f7f7f7f;
const int maxn = 400400;
const int maxm = 220;

int cnt, dhead[maxn];
int cur[maxn], dd[maxn];
Edge dedge[maxn<<1];
// bool vis[maxn]; // 记录经过的点
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) {
    // memset(vis, 0, sizeof(vis));
    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;
                // vis[dedge[i].v] = 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;
}

const int dx[11] = {-1,1,2,2,1,-1,-2,-2};
const int dy[11] = {-2,-2,-1,1,2,2,1,-1};
int n, m;
int mp[maxm][maxm];

inline bool bound(int x, int y) { return x >= 1 && x <= n && y >= 1 && y <= m;}
inline int id(int x, int y) { return (x - 1) * m + y; }

int main() {
    // freopen("in", "r", stdin);
    while(~scanf("%d%d",&n,&m)) {
        init();
        S = 0, T = 2 * id(n, m) + 1, N = T + 1;
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                adde(S, id(i,j), 1);
                adde(id(n,m)+id(i,j), T, 1);
                scanf("%d", &mp[i][j]);
            }
        }
        int tot = 0;
        for(int i = 1; i <= n; i++) {
            for(int j = 1; j <= m; j++) {
                if(mp[i][j] == 1) continue;
                tot++;
                for(int k = 0; k < 8; k++) {
                    int x = i + dx[k];
                    int y = j + dy[k];
                    if(!bound(x, y)) continue;
                    if(!mp[x][y]) adde(id(i,j), id(n,m)+id(x,y), inf);
                }
            }
        }
        printf("%d
", tot - dinic(S, T, N)/2);
    }
    return 0;
}