zoukankan      html  css  js  c++  java
  • 紫书 习题 11-9 UVa 12549 (二分图最小点覆盖)

    用到了二分图的一些性质, 最大匹配数=最小点覆盖

     貌似在白书上有讲

    还不是很懂, 自己看着别人的博客用网络流写了一遍

    反正以后学白书应该会系统学二分图的,紫书上没讲深。

    目前就这样吧。

    #include<cstdio>
    #include<vector>
    #include<cstring>
    #include<queue>
    #include<algorithm>
    #define REP(i, a, b) for(int i = (a); i < (b); i++)
    using namespace std;
    
    const int MAXN = 21234;
    const int MAXM = 112;
    struct Edge{ int from, to, cap, flow; };
    vector<Edge> edges;
    vector<int> g[MAXN];
    int cur[MAXN], h[MAXN];
    int n, m, s, t;
    int map[MAXM][MAXM], c[MAXM][MAXM], r[MAXM][MAXM];
    
    void AddEdge(int from, int to, int cap)
    {
    	edges.push_back(Edge{from, to, cap, 0});
    	edges.push_back(Edge{to, from, 0, 0});
    	g[from].push_back(edges.size() - 2);
    	g[to].push_back(edges.size() - 1);
    }
    
    bool bfs()
    {
    	queue<int> q;
    	q.push(s);
    	memset(h, 0, sizeof(h));
    	h[s] = 1;
    	
    	while(!q.empty())
    	{
    		int x = q.front(); q.pop();
    		REP(i, 0, g[x].size())
    		{
    			Edge& e = edges[g[x][i]];
    			if(e.cap > e.flow && !h[e.to])
    			{
    				h[e.to] = h[x] + 1;
    				q.push(e.to);
    			}
    		}
    	}
    	
    	return h[t];
    }
    
    int dfs(int x, int a)
    {
    	if(x == t || a == 0) return a;
    	int flow = 0, f;
    	for(int& i = cur[x]; i < g[x].size(); i++)
    	{
    		Edge& e = edges[g[x][i]];
    		if(h[x] + 1 == h[e.to] && (f = dfs(e.to, min(a, e.cap - e.flow))) > 0)
    		{
    			e.flow += f;
    			edges[g[x][i] ^ 1].flow -= f;
    			flow += f;
    			if((a -= f) == 0) break;
    		}
    	}
    	return flow;
    }
    
    int maxflow()
    {
    	int flow = 0;
    	while(bfs())
    	{
    		memset(cur, 0, sizeof(cur));
    		flow += dfs(s, 1e9);
    	}
    	return flow;
    }
    
    void make_edges()
    {
    	int cnt = 0, tmp;
    	REP(i, 0, n)
    	{
    		bool ok = true;
    		REP(j, 0, m)
    		{
    			if(map[i][j] == 1)
    			{
    				if(ok) cnt++;
    				r[i][j] = cnt;
    				ok = false;
    			}
    			else if(map[i][j] == 2) ok = true;
    		}
    	}
    	tmp = cnt;
    	
    	REP(j, 0, m)
    	{
    		bool ok = true;
    		REP(i, 0, n)
    		{
    			if(map[i][j] == 1)
    			{
    				if(ok) cnt++;
    				c[i][j] = cnt;
    				ok = false;
    			}
    			else if(map[i][j] == 2) ok = true;
    		}
    	}
    	
    	REP(i, 1, cnt + 5) g[i].clear();
    	s = cnt + 3, t = s + 1;
    	REP(i, 1, tmp + 1) AddEdge(s, i, 1);
    	REP(i, tmp + 1, cnt + 1) AddEdge(i, t, 1);
    	REP(i, 0, n)
    		REP(j, 0, m)
    			if(map[i][j] == 1)
    				AddEdge(r[i][j], c[i][j], 1);
    }
    
    int main()
    {
    	int T;
    	scanf("%d", &T);
    	
    	while(T--)
    	{
    		scanf("%d%d", &n, &m);
    		edges.clear();
    		memset(map, 0, sizeof(map));
    		memset(c, 0, sizeof(c));
    		memset(r, 0, sizeof(r));
    		
    		int tmp, x, y;
    		scanf("%d", &tmp);
    		while(tmp--)
    		{
    			scanf("%d%d", &x, &y); 
    			x--; y--;
    			map[x][y] = 1;
    		}
    		scanf("%d", &tmp);
    		while(tmp--)
    		{
    			scanf("%d%d", &x, &y);
    			x--; y--;
    			map[x][y] = 2;
    		}
    		
    		make_edges();
    		printf("%d
    ", maxflow());
    	}
    
    	return 0;
    }

  • 相关阅读:
    计算机考研真题 浮点数加法
    计算机考研复试真题 整数奇偶排序
    计算机考研复试 A+B
    计算机考研复试真题 整数拆分
    计算机考研复试真题 众数
    1121 Damn Single (25 分)
    1112 Stucked Keyboard (20 分)
    1117 Eddington Number (25 分)
    1005 继续(3n+1)猜想 (25 分)
    1047 编程团体赛 (20 分)
  • 原文地址:https://www.cnblogs.com/sugewud/p/9819524.html
Copyright © 2011-2022 走看看