[ZJOI2009]假期的宿舍

嘟嘟嘟

这几天刷最大流和最小割，正好某谷推荐了这道题，就顺便AC了一下。

好水。

建立源点和汇点，对于每一个能提供床位的同学，就从源点向该同学连一条容量为1的边；对于每一个需要床位的同学，就向汇点连一条容量为1的边。因此将每一个同学拆点。

接下来考虑同学之间的关系：如果x和y互相认识，就连一条(x, y +n)，(x +n, y)容量为1的边。

然后跑最大流，看总流量是否等于需要床位的人数。

#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 = 55;
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, t, tot = 0;
bool in[maxn];

struct Edge
{
    int from, to, cap, flow;
};
vector<Edge> edges;
vector<int> G[maxn << 1];
void addEdge(int from, int to)
{
    edges.push_back((Edge){from, to, 1, 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 << 1];
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 << 1];
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 init(int n)
{
    edges.clear();
    for(int i = 0; i <= n; ++i) G[i].clear();
    Mem(in, 0);
    tot = 0;
}

int main()
{
    int T = read();
    while(T--)
    {
        n = read(); t = (n << 1) + 1;
        init(t);
        for(int i = 1; i <= n; ++i) 
        {
            in[i] = (bool)read();
            if(in[i]) addEdge(0, i);
            else addEdge(i + n, t), tot++;
        }
        for(int i = 1; i <= n; ++i)
        {
            int x = read();
            if(in[i] && !x) addEdge(i + n, t), tot++;
        }
        for(int i = 1; i <= n; ++i)
            for(int j = 1; j <= n; ++j)
            {
                int x = read();
                if(x || i == j) addEdge(i, j + n), addEdge(j, i + n);
            }
        printf("%s
", maxflow() == tot ? "^_^" : "T_T");
    }
    return 0;
}