I - I WIN

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87954#problem/I

Description

Given an n × m rectangular tile with each square marked with one of the letters W, I, and N, find the maximal number of triominoes that can be cut from this tile such that the triomino has W and N on the ends and I in the middle (that is, it spells WIN in some order). Of course the only possible triominoes are the one with three squares in a straight line and the L-shaped ones, and the triominoes can't overlap.

Input

First line contains two integers n and m with 1 ≤ m, n ≤ 22. The next n lines contain m characters each (only the letters W, I and N).

Output

Output a single integer: the maximum number of nonoverlapping WIN-triominoes.

Sample Input

4 4

WIIW

NNNN

IINN

WWWI

Sample Output

5

HINT

 

题意

一个n*m的大矩形，问你最多切下多少只含有win的块数

题解：

网络流,s-1-w-1-i-1-i-1-n-t，这样建图

注意w连i的时候，i也要再另外开一个点连，然后连n

不然两个w连同一个i，两个n连同一个i的时候，会出现问题

代码：

#include 
     
      #include 
      
       #include 
       
        #include 
        
         using namespace std;namespace NetFlow{    const int MAXN=100000,MAXM=500000,inf=1e9;    struct Edge    {        int v,c,f,nx;        Edge() {}        Edge(int v,int c,int f,int nx):v(v),c(c),f(f),nx(nx) {}    } E[MAXM];    int G[MAXN],cur[MAXN],pre[MAXN],dis[MAXN],gap[MAXN],N,sz;    void init(int _n)    {        N=_n,sz=0; memset(G,-1,sizeof(G[0])*N);    }    void link(int u,int v,int c)    {        E[sz]=Edge(v,c,0,G[u]); G[u]=sz++;        E[sz]=Edge(u,0,0,G[v]); G[v]=sz++;    }    int ISAP(int S,int T)    {
    //S -> T        int maxflow=0,aug=inf,flag=false,u,v;        for (int i=0;i
         
          E[it].f&&dis[u]==dis[v=E[it].v]+1)                {                    if (aug>E[it].c-E[it].f) aug=E[it].c-E[it].f;                    pre[v]=u,u=v; flag=true;                    if (u==T)                    {                        for (maxflow+=aug;u!=S;)                        {                            E[cur[u=pre[u]]].f+=aug;                            E[cur[u]^1].f-=aug;                        }                        aug=inf;                    }                    break;                }            }            if (flag) continue;            int mx=N;            for (int it=G[u];~it;it=E[it].nx)            {                if (E[it].c>E[it].f&&dis[E[it].v]
          
           E[it].f) { dis[v]=dis[u]+1; Q[t++]=v; } } } return dis[T]!=-1; } int dfs(int u,int T,int low) { if (u==T) return low; int ret=0,tmp,v; for (int &it=cur[u];~it&&ret
           
            E[it].f) { if (tmp=dfs(v,T,min(low-ret,E[it].c-E[it].f))) { ret+=tmp; E[it].f+=tmp; E[it^1].f-=tmp; } } } if (!ret) dis[u]=-1; return ret; } int dinic(int S,int T) { int maxflow=0,tmp; while (bfs(S,T)) { memcpy(cur,G,sizeof(G[0])*N); while (tmp=dfs(S,T,inf)) maxflow+=tmp; } return maxflow; }}using namespace NetFlow;char s[40][40];int dx[4]={ 1,-1,0,0};int dy[4]={ 0,0,1,-1};int nn,mm;int judge(int x,int y){ if(x<=0||x>nn) return 0; if(y<=0||y>mm) return 0; return 1;}int main(){ //freopen("test.txt","r",stdin); scanf("%d%d",&nn,&mm); init(15000); for(int i=1;i<=nn;i++) scanf("%s",s[i]+1); for(int i=1;i<=nn;i++) { for(int j=1;j<=mm;j++) { if(s[i][j]=='I') { link(i+j*30,i+j*30+1000,1); } } } for(int i=1;i<=nn;i++) { for(int j=1;j<=mm;j++) { if(s[i][j]=='W') { link(4000,i+j*30,1); for(int k=0;k<4;k++) { if(judge(i+dx[k],j+dy[k])) { if(s[i+dx[k]][j+dy[k]]=='I') { int ii=i+dx[k],jj=j+dy[k]; link(i+j*30,ii+jj*30,1); } } } } if(s[i][j]=='I') { for(int k=0;k<4;k++) { if(judge(i+dx[k],j+dy[k])) { if(s[i+dx[k]][j+dy[k]]=='N') { int ii=i+dx[k],jj=j+dy[k]; link(i+j*30+1000,ii+jj*30,1); } } } } if(s[i][j]=='N') { link(i+j*30,4001,1); } } } printf("%d\n",dinic(4000,4001)); return 0;}