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POJ 2112 —— Optimal Milking 二分+Floyd+最大流

时间:2016-05-07 10:18:18      阅读:116      评论:0      收藏:0      [点我收藏+]

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原题:http://poj.org/problem?id=2112


#include<cstdio>
#include<cstring>
#include<string>
#include<queue>
#include<vector>
#include<algorithm>
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 250;
int dis[maxn][maxn];
int k, c, m;
int num_nodes;

struct Edge  
{  
    int from, to, flow, cap;  
}edge[maxn*maxn*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);  
}

void Floyd()
{
	for(int t = 1;t<=k+c;t++)
	{
		for(int i = 1;i<=k+c;i++)
		{
			for(int j = 1;j<=k+c;j++)
			dis[i][j] = min(dis[i][j], dis[i][t]+dis[t][j]);
		}
	}
}

int main()
{
	while(~scanf("%d%d%d", &k, &c, &m))
	{
		for(int i = 1;i<=k+c;i++)
		{
			for(int j = 1;j<=k+c;j++)
			{
				scanf("%d", &dis[i][j]);
				if(i != j && dis[i][j] == 0)
					dis[i][j] = inf;
			}
		}
		Floyd();
		int l = 0, r = 10000;		
		int s = 0, t = k+c+1;
		num_nodes = t+1;
		int ans = inf;		
		while(l <= r)
		{
			int mid = (l+r)/2;
			init();
			for(int i = k+1;i<=k+c;i++)
			{
				for(int j = 1;j<=k;j++)
				{
					if(dis[i][j] <= mid)
					add(i, j, 1);
				}
			}
			for(int i = k+1;i<=k+c;i++)
				add(s, i, 1);
			for(int i = 1;i<=k;i++)
				add(i, t, m);
			int flow = Maxflow(s, t);
			if(flow != c)
				l = mid + 1;
			else
			{
				ans = min(ans, mid);
				r = mid - 1;
			}
		}
		printf("%d\n", ans);
	}
	return 0;
}


POJ 2112 —— Optimal Milking 二分+Floyd+最大流

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原文地址:http://blog.csdn.net/l_avender/article/details/51332309

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