bzoj1305 [CQOI2009]dance跳舞

传送门：http://www.lydsy.com/JudgeOnline/problem.php?id=1305

【题解】

把男孩看成点，女孩看成点，一眼就知道是个匹配模型。

把男孩拆成B1,B2，女孩拆成G1,G2

$B1_i ightarrow B2_i, [k]$

$G2_i ightarrow G1_i, [k]$

表示至多跟k个不喜欢的人跳舞。

如果$(i,j)$喜欢，那么$B1_i ightarrow G1_j ,[1]$

否则$B2_i ightarrow G2_j, [1]$

仔细想想发现十分科学。

二分答案$x$，建边$S ightarrow B1_i ,[x]$，$G1_i ightarrow T ,[x]$。

一个j打成i的傻逼错误检查了半天。。

# include <queue>
# include <stdio.h>
# include <assert.h> 
# include <string.h>
# include <iostream>
# include <algorithm>
// # include <bits/stdc++.h>

using namespace std;

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
const int M = 5e5 + 10, N = 60;
const int mod = 1e9+7, inf = 1e9;

int n, K; 
char mp[N][N]; 

# define B1(x) (x)
# define B2(x) ((x) + n)
# define G1(x) ((x) + n + n)
# define G2(x) ((x) + n + n + n) 

int S, T, id[M], idn;
namespace MF {
    int head[N * 5], nxt[M], to[M], flow[M], tot = 1;
    inline void add(int u, int v, int fl) {
        ++tot; nxt[tot] = head[u]; head[u] = tot; to[tot] = v; flow[tot] = fl;
    }
    inline void adde(int u, int v, int fl) {
        add(u, v, fl), add(v, u, 0);
    }
    queue<int> q; int c[N * 5], cur[N * 5];
    inline bool bfs() {
         while(!q.empty()) q.pop();
        for (int i=1; i<=T; ++i) c[i] = -1;
        c[S] = 0; q.push(S);
        while(!q.empty()) {
            int top = q.front(); q.pop();
            for (int i=head[top]; i; i=nxt[i]) {
                if(c[to[i]] != -1 || !flow[i]) continue;
                c[to[i]] = c[top] + 1;
                q.push(to[i]);
                if(to[i] == T) return 1;
            }
        }
        return 0;
    }
    inline int dfs(int x, int low) {
        if(x == T) return low;
        int r = low, fl;
        for (int i=cur[x]; i; i=nxt[i]) {
             if(c[to[i]] != c[x] + 1 || !flow[i]) continue;
            fl = dfs(to[i], min(r, flow[i]));
            flow[i] -= fl; r -= fl; flow[i^1] += fl;
            if(flow[i] > 0) cur[x] = i;
            if(!r) return low;
        }
        if(low == r) c[x] = -1;
        return low-r;
    }
    inline int MF() {
        int ret = 0;
        while(bfs()) {
            for (int i=1; i<=T; ++i) cur[i] = head[i];
            ret += dfs(S, inf);
        }
        return ret;
    }
    inline int main(int x) {
        for (int i=1; i<=idn; ++i) {
            int s = id[i]^1;
            flow[s] = x;
        }
        int ret = MF();
        for (int i=2; i<=tot; i+=2) {
            flow[i] += flow[i^1];
            flow[i^1] = 0;
        }
        return ret == x * n;
    }
}


int main() {
    cin >> n >> K; S = n + n + n + n + 1, T = S + 1;
    for (int i=1; i<=n; ++i) {
        scanf("%s", mp[i] + 1);
        for (int j=1; j<=n; ++j)
            if(mp[i][j] == 'Y') MF :: adde(B1(i), G1(j), 1);
            else MF :: adde(B2(i), G2(j), 1);
    }
    for (int i=1; i<=n; ++i) {
        MF :: adde(B1(i), B2(i), K);
        MF :: adde(G2(i), G1(i), K);
        MF :: adde(S, B1(i), inf);
        id[++idn] = MF :: tot;
        MF :: adde(G1(i), T, inf);
        id[++idn] = MF :: tot; 
    } 
    
    int l = 0, r = n, mid;
    while(1) {
        if(r - l <= 3) {
            for (int i=r; i>=l; --i)
                if(MF :: main(i)) {
                    cout << i;
                    return 0;
                }
            break;
        }
        mid = l+r>>1;
        if(MF :: main(mid)) l = mid;
        else r = mid;
    }
    assert(0); 
    return 0;
}