标签：acm   网络流   图论   algorithm   

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=504

Crimewave

Nieuw Knollendam is a very modern town. This becomes clear already when looking at the layout of its map, which is just a rectangular grid of streets and avenues. Being an important trade centre, Nieuw Knollendam also has a lot of banks. Almost on every crossing a bank is found (although there are never two banks at the same crossing). Unfortunately this has attracted a lot of criminals. Bank hold-ups are quite common, and often on one day several banks are robbed. This has grown into a problem, not only to the banks, but to the criminals as well. After robbing a bank the robber tries to leave the town as soon as possible, most of the times chased at high speed by the police. Sometimes two running criminals pass the same crossing, causing several risks: collisions, crowds of police at one place and a larger risk to be caught.

To prevent these unpleasant situations the robbers agreed to consult together. Every Saturday night they meet and make a schedule for the week to come: who is going to rob which bank on which day? For every day they try to plan the get-away routes, such that no two routes use the same crossing. Sometimes they do not succeed in planning the routes according to this condition, although they believe that such a planning should exist.

Given a grid of  and the crossings where the banks to be robbed are located, find out whether or not it is possible to plan a get-away route from every robbed bank to the city-bounds, without using a crossing more than once.

Input 

• The first line of every problem contains the number s of streets ( ), followed by the number a of avenues ( ), followed by the number b () of banks to be robbed.

• Then b lines follow, each containing the location of a bank in the form of two numbers x (the number of the street) and y (the number of the avenue). Evidently  and .

Output 

Sample Input 

2
6 6 10
4 1
3 2
4 2
5 2
3 4
4 4
5 4
3 6
4 6
5 6
5 5 5
3 2
2 3
3 3
4 3
3 4

Sample Output 

possible
not possible

题意：

有若干罪犯抢银行，要求逃出地图时他们的路线不相交，求是否能达到上述要求。

分析：

路线不相交即每个点每条边只能用一次，即容量为1,点上的流量限制拆点即可。源点连向罪犯所在位置的入点，最外一圈的出点连向汇点，满流即可能。

/*
 *
 *	Author	:	fcbruce
 *
 *	Date	:	2014-09-04 21:26:22 
 *
 */
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10

#ifdef _WIN32
	#define lld "%I64d"
#else
	#define lld "%lld"
#endif

#define maxm 2333333
#define maxn 8964

using namespace std;

int fir[maxn];
int u[maxm],v[maxm],cap[maxm],flow[maxm],nex[maxm];
int e_max;
int iter[maxn],q[maxn],lv[maxn];

void add_edge(int _u,int _v,int _w)
{
    int e;
    e=e_max++;
    u[e]=_u;v[e]=_v;cap[e]=_w;
    nex[e]=fir[u[e]];fir[u[e]]=e;
    e=e_max++;
    u[e]=_v;v[e]=_u;cap[e]=0;
    nex[e]=fir[u[e]];fir[u[e]]=e;
}

void dinic_bfs(int s)
{
    int f,r;
    memset(lv,-1,sizeof lv);
    q[f=r=0]=s;
    lv[s]=0;
    while(f<=r)
    {
        int x=q[f++];
        for (int e=fir[x];~e;e=nex[e])
        {
            if (cap[e]>flow[e] && lv[v[e]]<0)
            {
                lv[v[e]]=lv[u[e]]+1;
                q[++r]=v[e];
            }
        }
    }
}

int dinic_dfs(int _u,int t,int _f)
{
    if (_u==t)  return _f;
    for (int &e=iter[_u];~e;e=nex[e])
    {
        if (cap[e]>flow[e] && lv[_u]<lv[v[e]])
        {
            int _d=dinic_dfs(v[e],t,min(_f,cap[e]-flow[e]));
            if (_d>0)
            {
                flow[e]+=_d;
                flow[e^1]-=_d;
                return _d;
            }
        }
    }

    return 0;
}

int max_flow(int s,int t)
{

    memset(flow,0,sizeof flow);
    int total_flow=0;

    for (;;)
    {
        dinic_bfs(s);
        if (lv[t]<0)    break;
        memcpy(iter,fir,sizeof iter);
        int _f;

        while ((_f=dinic_dfs(s,t,INF))>0)
            total_flow+=_f;
    }

    return total_flow;
}

int main()
{
	#ifdef FCBRUCE
		freopen("/home/fcbruce/code/t","r",stdin);
	#endif // FCBRUCE
	
	int T_T;
	
	scanf( "%d",&T_T);
	
	while (T_T--)
	{
		int n,m,s=0,t=8963;
		scanf( "%d%d",&n,&m);
		
		e_max=0;
		memset(fir,-1,sizeof fir);
		
		for (int i=1;i<=n;i++)
			for (int j=1;j<=m;j++)
			{
				add_edge(i*m+j+n*m,i*m+j,1);
				
				if (i==n || j==m) continue;
				
				add_edge(i*m+j,(i+1)*m+j+n*m,1);
				add_edge(i*m+j,i*m+j+1+n*m,1);
				
				add_edge((i+1)*m+j,i*m+j+n*m,1);
				add_edge(i*m+j+1,i*m+j+n*m,1);
			}
			
		for (int i=1;i<=m;i++)
		{
			add_edge(1*m+i,t,1);
			add_edge(n*m+i,t,1);
		}
		
		for (int i=2;i<n;i++)
		{
			add_edge(i*m+1,t,1);
			add_edge(i*m+m,t,1);
		}
		
		int p;
		scanf( "%d",&p);
		for (int i=0,x,y;i<p;i++)
		{
			scanf( "%d%d",&x,&y);
			add_edge(s,x*m+y+n*m,1);
		}
		
		if (max_flow(s,t)==p)
			puts( "possible");
		else
			puts( "not possible");
	}
	
	return 0;
}

UVA 563 Crimewave (最大流，拆点)

标签：acm   网络流   图论   algorithm   

原文地址：http://blog.csdn.net/u012965890/article/details/39060523