POJ 1087 —— A Plug for UNIX

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原题：http://poj.org/problem?id=1087

题意：n个插座，m个电器及其对应的插座，k种转化器，转换器（u，v）表示可以把原本需要u插座的电器转接到v插座上，问最少有多少设备没有插座用，每种转换器数量不限;；

#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#include<map>
#include<algorithm>
#define inf 1e9
using namespace std;
const int maxn = 1500;
const int maxm = 5500;
int n, m, k;
int num_nodes;
int a[30];
map<string, int>mp;

struct Edge  
{  
    int from, to, flow, cap;  
}edge[maxm*2];  
  
vector<int>G[maxn]; 
int edgenum;  
void add(int u, int v, int c) 
{  
    edge[edgenum].from = u;  
    edge[edgenum].to = v;  
    edge[edgenum].flow = 0;  
    edge[edgenum].cap = c;  
    edgenum++;  
      
    edge[edgenum].from = v;  
    edge[edgenum].to = u;  
    edge[edgenum].flow = 0;  
    edge[edgenum].cap = 0;  
    edgenum++;  
      
    G[u].push_back(edgenum-2);  
    G[v].push_back(edgenum-1);  
}  
  
int deep[maxn];  
bool vis[maxn];  
void BFS(int s, int t)  
{  
    queue<int>Q;  
    memset(vis, false, sizeof vis);  
    Q.push(t);  
    vis[t] = true;  
    deep[t] = 0;  
    while(!Q.empty())  
    {  
        int now = Q.front();  
        Q.pop();  
        for(int i = 0;i<(int)G[now].size();i++)  
        {  
            int v = edge[G[now][i]].to;  
            if(!vis[v]) 
			{  
                deep[v] = deep[now] + 1;  
                vis[v] = true;  
                Q.push(v);  
            }  
        }  
    }  
}  
  
int gap[maxn]; 
int cur[maxn]; 
int front[maxn];  
int Augment(int s, int t)  
{  
    int minflow = inf;  
    int begin = t;  
    while(begin != s)  
    {  
        Edge& e = edge[front[begin]];  
        minflow = min(minflow, e.cap - e.flow);  
        begin = e.from;  
    }  
      
    begin = t;  
    while(begin != s)  
    {  
        edge[front[begin]].flow += minflow;  
        edge[front[begin]^1].flow -= minflow;  
        begin = edge[front[begin]].from;  
    }  
    return minflow;  
}  
  
int Maxflow(int s, int t)  
{  
    int flow = 0;  
    BFS(s, t);  
    memset(gap, 0, sizeof gap);  
    memset(cur, 0, sizeof cur);  
    for(int i = 0;i<num_nodes;i++)  gap[deep[i]]++;  
    int begin = s;  
    while(deep[s] < num_nodes)  
    {  
        if(begin == t) 
		{  
            flow += Augment(s, t);  
            begin = s;  
        }  
        bool flag = false;  
        for(int i = cur[begin];i<(int)G[begin].size();i++)  
        {  
            Edge& e = edge[G[begin][i]];  
            if(e.cap > e.flow && deep[begin] == deep[e.to] + 1) 
			{  
                front[e.to] = G[begin][i];  
                cur[begin] = i;  
                flag = true;  
                begin = e.to;  
                break;  
            }  
        }  
        if(!flag)  
        {  
            int k = num_nodes-1;  
            for(int i = 0;i<(int)G[begin].size();i++) 
			{  
                Edge& e = edge[G[begin][i]];  
                if(e.cap > e.flow)  
                    k = min(k, deep[e.to]);  
            }  
            if(--gap[deep[begin]] == 0) break; 
            gap[deep[begin] = k+1]++;  
            cur[begin] = 0;  
            if(begin != s)   
                begin = edge[front[begin]].from;   
        }  
    }  
    return flow;  
}  
  
void init() 
{  
    for(int i = 0;i<num_nodes+2;i++) G[i].clear();  
    edgenum = 0;  
    memset(deep, 0, sizeof deep);  
}

int cnt = 0;
int Id(char s[])
{
	if(mp.find(s) == mp.end())
		mp[s] = ++cnt;
	return mp[s];
}

int main()
{
	int s = 0, t = 400;
	num_nodes = t+1;
	init();
	scanf("%d", &n);
	char c[5];
	mp.clear();	
	for(int i = 1;i<=n;i++)
	{
		scanf("%s", c);	
		add(Id(c), t, 1);	
	}
	scanf("%d", &m);	
	char thing[30];
	for(int i = 1;i<=m;i++)
	{		
		scanf("%s%s", thing, c);
		add(s, i+n, 1);
		add(i+n, Id(c), 1);
	}
	scanf("%d", &k);
	char u[5], v[5];
	for(int i = 1;i<=k;i++)
	{
		
		scanf("%s%s", u, v);
		add(Id(u), Id(v), inf);
	}
	int flow = Maxflow(s, t);
	printf("%d\n", m-flow);
	return 0;
}

POJ 1087 —— A Plug for UNIX

标签：

原文地址：http://blog.csdn.net/l_avender/article/details/51357168