[CQOI2009]跳舞

嘟嘟嘟

因为如果能跳x支舞曲，那么一定能跳y(y < x)支舞曲，满足单调性。因此可二分舞曲。

那么网络流建图就明白了：把每一个男生拆成3个点，分别是男生总，男生喜欢的，男生不喜欢的。然后从源点向男生总连一条容量为x的边，从男生总向男生喜欢的连一条INF的边，向男生不喜欢的连一条k的边。女生同理。

然后对于一对互相喜欢的男女生，就从男生喜欢的向女生喜欢的连一条1的边；对于不喜欢的男女生，就从男生不喜欢的向女生不喜欢的连一条1的边。跑一遍最大流，如果总流量等于x * n，就说明可以跳x支舞曲，向上二分。

#include<cstdio>
#include<iostream>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<vector>
#include<stack>
#include<queue>
using namespace std;
#define enter puts("") 
#define space putchar(' ')
#define Mem(a, x) memset(a, x, sizeof(a))
#define rg register
typedef long long ll;
typedef double db;
const int INF = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 50;
inline ll read()
{
    ll ans = 0;
    char ch = getchar(), last = ' ';
    while(!isdigit(ch)) {last = ch; ch = getchar();}
    while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
    if(last == '-') ans = -ans;
    return ans;
}
inline void write(ll x)
{
    if(x < 0) x = -x, putchar('-');
    if(x >= 10) write(x / 10);
    putchar(x % 10 + '0');
}

int n, k, t;
char a[maxn + 5][maxn + 5];

struct Edge
{
    int from, to, cap, flow;
};
vector<Edge> edges;
vector<int> G[maxn * 6 + 5];
void addEdge(int from, int to, int w)
{
    edges.push_back((Edge){from, to, w, 0});
    edges.push_back((Edge){to, from, 0, 0});
    int sz = edges.size();
    G[from].push_back(sz - 2);
    G[to].push_back(sz - 1);
}

int dis[maxn * 6 + 5];
bool bfs()
{
    Mem(dis, 0); dis[0] = 1;
    queue<int> q; q.push(0);
    while(!q.empty())
    {
        int now = q.front(); q.pop();
        for(int i = 0; i < (int)G[now].size(); ++i)
        {
            Edge& e = edges[G[now][i]];
            if(!dis[e.to] && e.cap > e.flow)
            {
                dis[e.to] = dis[now] + 1;
                q.push(e.to);    
            }
        }
    }
    return dis[t];
}
int cur[maxn * 6 + 5];
int dfs(int now, int res)
{
    if(now == t || res == 0) return res;
    int flow = 0, f;
    for(int& i = cur[now]; i < (int)G[now].size(); ++i)
    {
        Edge& e = edges[G[now][i]];
        if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(res, e.cap - e.flow))) > 0)
        {
            e.flow += f;
            edges[G[now][i] ^ 1].flow -= f;
            flow += f; res -= f;
            if(res == 0) break;
        }
    }
    return flow;
}

int maxflow()
{
    int flow = 0;
    while(bfs())
    {
        Mem(cur, 0);
        flow += dfs(0, INF);
    }
    return flow;
}

void build_Gra(int x)        //暴力重构 
{
    edges.clear();            
    for(int i = 0; i <= t; ++i) G[i].clear();
    for(int i = 1; i <= n; ++i)
    {
        addEdge(0, i, x);
        addEdge(i, n + i, INF); 
        addEdge(i, (n << 1) + i, k);
        addEdge(n * 5 + i, t, x); 
        addEdge((n << 1) + n + i, n * 5 + i, INF);
        addEdge((n << 2) + i, n * 5 + i, k);
        for(int j = 1; j <= n; ++j) 
        {
            if(a[i][j] == 'Y') addEdge(n + i, (n << 1) + n + j, 1);
            else addEdge((n << 1) + i, (n << 2) + j, 1); 
        }
    }
}

int main()
{
    n = read(); k = read();
    t = (n << 2) + (n << 1) + 1;
    for(int i = 1; i <= n; ++i) scanf("%s", a[i] + 1);
    int L = 0, R = n;
    while(L < R)
    {
        int mid = (L + R + 1) >> 1;
        build_Gra(mid);
        if(maxflow() == mid * n) L = mid;
        else R = mid - 1;
    }
    write(L); enter;
    return 0;
}