Ultraviolet A 563 crimewave (maximum stream, split point)

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 center, 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 shocould 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 numberSOf streets (), followed by the numberAOf avenues (), followed by the numberB() Of banks to be robbed.

• ThenBLines follow, each containing the location of a bank in the form of two numbersX(The number of the street) andY(The number of the Avenue). Evidently and.

Output the output file consists of P lines. Each line contains the text PossibleOr Not possible. If it is possible to plan non-crossing get-away routes, this line shoshould contain the word: Possible. If this is not possible, the line shoshould contain the words Not possible.

Sample Input

26 6 104 13 24 25 23 44 45 43 64 65 65 5 53 22 33 34 33 4

Sample output

possiblenot possible

Miguel A. Revilla
1998

Question:

Several criminals have snatched the bank and demanded that their routes do not match when they escaped from the map.

Analysis:

If the route is not intersecting, each edge of each vertex can be used only once, that is, the capacity is 1, and the traffic limit on the vertex can be split. The source point is connected to the entry point where the criminal is located, and the outgoing point in the outermost circle is connected to the sink point. full stream is possible.

/* * *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 8964using 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 FCBRUCEfreopen("/home/fcbruce/code/t","r",stdin);#endif // FCBRUCEint 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");elseputs( "not possible");}return 0;}

Ultraviolet A 563 crimewave (maximum stream, split point)