<center>
| 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 the input contains the number of problems p to be solved.- The first line of every problem contains the number s of streets (
), followed by the numbera 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
The output file consists of p lines. Each line contains the text <tt>possible</tt> or <tt>not possible</tt> . If it is possible to plan non-crossing get-away routes, this line should contain the word: <tt>possible</tt> . If this is not possible, the line should contain the words <tt>not possible</tt> .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：街道相互垂直，每个交点有一家银行。许多强盗某一时刻抢劫不同的银行，离开网格的时候路线不能相交，问是否存在此路线
做法：看下面的代码吧……自己的错误是忘记了长方形的另外两条边==太不应该了
/*************
uva 563
2016-08-30 10:54:18
0 0
C++ 5.3.0
2725
************/
#include <stdio.h>
#include<cstring>
#include <iostream>
#include<vector>
#include<algorithm>
#include<cstring>
using namespace std;
const int oo=0x3f3f3f3f;
const int mm=900000;
const int mn=900000;
int node ,scr,dest,edge;
int ver[mm],flow[mm],Next[mm];
int head[mn],work[mn],dis[mn],q[mn];
void prepare(int _node,int _scr,int _dest)
{
node=_node,scr=_scr,dest=_dest;
for(int i=0; i<node; ++i)
head[i]=-1;
edge=0;
}
void addedge(int u,int v,int c)
{
ver[edge]=v,flow[edge]=c,Next[edge]=head[u],head[u]=edge++;
ver[edge]=u,flow[edge]=0,Next[edge]=head[v],head[v]=edge++;
}
bool Dinic_bfs()
{
int i,u,v,l,r=0;
for(i=0; i<node; i++)
dis[i]=-1;
dis[q[r++]=scr]=0;
for(l=0; l<r; ++l)
{
for(i=head[u=q[l]]; i>=0; i=Next[i])
{
if(flow[i]&&dis[v=ver[i]]<0)
{
dis[q[r++]=v]=dis[u]+1;
if(v==dest)
return 1;
}
}
}
return 0;
}
int Dinic_dfs(int u,int exp)
{
if(u==dest)
return exp;
for(int &i=work[u],v,tmp; i>=0; i=Next[i])
if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0)
{
flow[i]-=tmp;
flow[i^1]+=tmp;
return tmp;
}
return 0;
}
int Dinic_flow()
{
int i,ret=0,delta;
while(Dinic_bfs())
{
for(i=0; i<node; i++)
work[i]=head[i];
while(delta=Dinic_dfs(scr,oo))
ret+=delta;
}
return ret;
}
int n,m,t,Q;
int main()
{
// freopen("cin.txt","r",stdin);
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d",&n,&m,&Q);
prepare(n*m*2+2,0,n*m*2+1);
for(int i=1;i<=Q;i++)
{
int a,b;
scanf("%d%d",&a,&b);
addedge(0,a+(b-1)*n,1);
}
for(int i=1;i<=n*m;i++)addedge(i,i+n*m,1);
//
for(int i=1+n*m;i<=n+n*m;i++)addedge(i,dest,1);
for(int i=2;i<=n;i++)addedge((m-1)*n+i+n*m,dest,1);
for(int i=n+1+n*m;i<=n*m*2;i+=n)addedge(i,dest,1);
for(int i=2;i<m;i++)addedge(i*n+n*m,dest,1);
//
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
//(i,j)=>(i-1,j) (i+1,j) (i,j+1) (i,j-1)
if(i>1) addedge(i+(j-1)*n+n*m,i-1+(j-1)*n,1);
if(i<n) addedge(i+(j-1)*n+n*m,i+1+(j-1)*n,1);
if(j>1) addedge(i+(j-1)*n+n*m,i+(j-2)*n,1);
if(j<m) addedge(i+(j-1)*n+n*m,i+(j)*n,1);
}
}
//printf("%d\n",);
if(Q==Dinic_flow())puts("possible");
else puts("not possible");
}
return 0;
}
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