2012 If this is the end of the world how to do? I do not know how. But now scientists have found that some stars, who can live, but some people do not fit to live some of the planet. Now scientists want your help, is to determine
what all of people can live in these planets.
The last line has m digits, the ith digit ai indicates the ith planet can contain ai people most..
0 <= ai <= 100000
If you can output YES, otherwise output NO.
1 1 1 1 2 2 1 0 1 0 1 1
YES NO最大流
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;
const int inf = 0x3f3f3f3f;
const int mod=609929123;
const int N=100220;
int frist[N];
int current[N];
int dis[N];
char mapp[100][100];
char mapp1[100][100];
int mp[100][100];
int number[100][100];
int R,S,T,t,n,d,m,cnt;
struct node
{
int v,cap,next;//领接表
}es[4*N];
int addedge(int u,int v,int cap)
{
node e1={v,cap,frist[u]};
es[R]=e1;
frist[u]=R++;
node e2={u,0,frist[v]};
es[R]=e2;
frist[v]=R++;
}
int bfs()
{
queue<int>q;
q.push(S);
memset(dis,-1,sizeof(dis));
dis[S]=0;
while(!q.empty())
{
int h=q.front();
if(h==T) return 1;
q.pop();
for(int i=frist[h];i!=-1;i=es[i].next)
{
int temp=es[i].v;
if(dis[temp]==-1&&es[i].cap)
{
dis[temp]=dis[h]+1;
q.push(temp);
}
}
}
return 0;
}
int dinic(int x,int maxflow)
{
if(x==T) return maxflow;
int flow,f=0;
for(int &i=current[x];i!=-1;i=es[i].next)
{
int temp=es[i].v;
if(dis[temp]==dis[x]+1&&es[i].cap)
{
flow=dinic(temp,min(maxflow-f,es[i].cap));
es[i].cap-=flow;
es[i^1].cap+=flow;
f+=flow;
if(f==maxflow) break;
}
}
return f;
}
int DINIC()
{
int ans=0;
while(bfs())
{
int flow;
memcpy(current,&frist,sizeof(frist));//复制
while(flow=dinic(S,inf))
ans+=flow;
}
return ans;
}
int main()
{
scanf("%d",&t);
int o=0;
while(t--)
{
memset(frist,-1,sizeof(frist));
memset(mapp,0,sizeof(mapp));
memset(mapp1,0,sizeof(mapp1));
memset(mp,0,sizeof(mp));
memset(number,0,sizeof(number));
R=0,S=0;
cnt=0;
scanf("%d%d",&n,&d);
for(int i=1;i<=n;i++)
{
scanf("%s",mapp[i]+1);
m=strlen(mapp[i]+1);
for(int j=1;j<=m;j++)
{
mp[i][j]=mapp[i][j]-'0';
number[i][j]=++cnt;
}
}
T=2*cnt+1;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m;j++)
{
if(mp[i][j])
{
if(i<=d||j<=d||i+d>n||j+d>m)
addedge(number[i][j]+cnt,T,inf);
addedge(number[i][j],number[i][j]+cnt,mp[i][j]);
for(int k=1;k<=n;k++)
{
for(int p=1;p<=m;p++)
{
int dx=abs(i-k);
int dy=abs(j-p);
double mm=sqrt(dx*dx*1.0+dy*dy*1.0);
if(mm>d) continue;
addedge(number[i][j]+cnt,number[k][p],inf);
}
}
}
}
}
int ll=0;
for(int i=1;i<=n;i++)
{
scanf("%s",mapp1[i]+1);
for(int j=1;j<=m;j++)
{
if(mapp1[i][j]=='L')
{
ll++;
addedge(0,number[i][j],1);
}
}
}
ll-=DINIC();
if(ll==0)
printf("Case #%d: no lizard was left behind.\n",++o);
else if(ll==1)
printf("Case #%d: 1 lizard was left behind.\n",++o);
else
printf("Case #%d: %d lizards were left behind.\n",++o,ll);
}
}