[ZJOI2009]假期的宿舍

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。

  1 #include<cstdio>
  2 #include<iostream>
  3 #include<cmath>
  4 #include<algorithm>
  5 #include<cstring>
  6 #include<cstdlib>
  7 #include<cctype>
  8 #include<vector>
  9 #include<stack>
 10 #include<queue>
 11 using namespace std;
 12 #define enter puts("") 
 13 #define space putchar(' ')
 14 #define Mem(a, x) memset(a, x, sizeof(a))
 15 #define rg register
 16 typedef long long ll;
 17 typedef double db;
 18 const int INF = 0x3f3f3f3f;
 19 const db eps = 1e-8;
 20 const int maxn = 55;
 21 inline ll read()
 22 {
 23     ll ans = 0;
 24     char ch = getchar(), last = ' ';
 25     while(!isdigit(ch)) {last = ch; ch = getchar();}
 26     while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();}
 27     if(last == '-') ans = -ans;
 28     return ans;
 29 }
 30 inline void write(ll x)
 31 {
 32     if(x < 0) x = -x, putchar('-');
 33     if(x >= 10) write(x / 10);
 34     putchar(x % 10 + '0');
 35 }
 36 
 37 int n, t, tot = 0;
 38 bool in[maxn];
 39 
 40 struct Edge
 41 {
 42     int from, to, cap, flow;
 43 };
 44 vector<Edge> edges;
 45 vector<int> G[maxn << 1];
 46 void addEdge(int from, int to)
 47 {
 48     edges.push_back((Edge){from, to, 1, 0});
 49     edges.push_back((Edge){to, from, 0, 0});
 50     int sz = edges.size();
 51     G[from].push_back(sz - 2);
 52     G[to].push_back(sz - 1);
 53 }
 54 
 55 int dis[maxn << 1];
 56 bool bfs()
 57 {
 58     Mem(dis, 0); dis[0] = 1;
 59     queue<int> q; q.push(0);
 60     while(!q.empty())
 61     {
 62         int now = q.front(); q.pop();
 63         for(int i = 0; i < (int)G[now].size(); ++i)
 64         {
 65             Edge& e = edges[G[now][i]];
 66             if(!dis[e.to] && e.cap > e.flow)
 67             {
 68                 dis[e.to] = dis[now] + 1;
 69                 q.push(e.to);    
 70             }
 71         }
 72     }
 73     return dis[t];
 74 }
 75 int cur[maxn << 1];
 76 int dfs(int now, int res)
 77 {
 78     if(now == t || res == 0) return res;
 79     int flow = 0, f;
 80     for(int& i = cur[now]; i < (int)G[now].size(); ++i)
 81     {
 82         Edge& e = edges[G[now][i]];
 83         if(dis[e.to] == dis[now] + 1 && (f = dfs(e.to, min(res, e.cap - e.flow))) > 0)
 84         {
 85             e.flow += f;
 86             edges[G[now][i] ^ 1].flow -= f;
 87             flow += f; res -= f;
 88             if(res == 0) break;
 89         }
 90     }
 91     return flow;
 92 }
 93 
 94 int maxflow()
 95 {
 96     int flow = 0;
 97     while(bfs())
 98     {
 99         Mem(cur, 0);
100         flow += dfs(0, INF);
101     }
102     return flow;
103 }
104 
105 void init(int n)
106 {
107     edges.clear();
108     for(int i = 0; i <= n; ++i) G[i].clear();
109     Mem(in, 0);
110     tot = 0;
111 }
112 
113 int main()
114 {
115     int T = read();
116     while(T--)
117     {
118         n = read(); t = (n << 1) + 1;
119         init(t);
120         for(int i = 1; i <= n; ++i) 
121         {
122             in[i] = (bool)read();
123             if(in[i]) addEdge(0, i);
124             else addEdge(i + n, t), tot++;
125         }
126         for(int i = 1; i <= n; ++i)
127         {
128             int x = read();
129             if(in[i] && !x) addEdge(i + n, t), tot++;
130         }
131         for(int i = 1; i <= n; ++i)
132             for(int j = 1; j <= n; ++j)
133             {
134                 int x = read();
135                 if(x || i == j) addEdge(i, j + n), addEdge(j, i + n);
136             }
137         printf("%s
", maxflow() == tot ? "^_^" : "T_T");
138     }
139     return 0;
140 }

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