POJ2711 Leapin' Lizards（最大流）

比较形象的是地图每个点都拆成三个点，这三个点限制流量为0或1，于是再一分为二，这样每个点都被拆成6个点。。。

其实拆两个点，连容量为柱子高的边，这样就行了。。

这题我掉坑了，“1 lizard was left behind.”。。虽然样例都把一切都说了。。要注意细节。。

#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
#define INF (1<<30)
#define MAXN 888 
#define MAXM 888*888*2 

struct Edge{
    int v,cap,flow,next;
}edge[MAXM];
int vs,vt,NE,NV;
int head[MAXN];

void addEdge(int u,int v,int cap){
    edge[NE].v=v; edge[NE].cap=cap; edge[NE].flow=0;
    edge[NE].next=head[u]; head[u]=NE++;
    edge[NE].v=u; edge[NE].cap=0; edge[NE].flow=0;
    edge[NE].next=head[v]; head[v]=NE++;
}

int level[MAXN];
int gap[MAXN];
void bfs(){
    memset(level,-1,sizeof(level));
    memset(gap,0,sizeof(gap));
    level[vt]=0;
    gap[level[vt]]++;
    queue<int> que;
    que.push(vt);
    while(!que.empty()){
        int u=que.front(); que.pop();
        for(int i=head[u]; i!=-1; i=edge[i].next){
            int v=edge[i].v;
            if(level[v]!=-1) continue;
            level[v]=level[u]+1;
            gap[level[v]]++;
            que.push(v);
        }
    }
}

int pre[MAXN];
int cur[MAXN];
int ISAP(){
    bfs();
    memset(pre,-1,sizeof(pre));
    memcpy(cur,head,sizeof(head));
    int u=pre[vs]=vs,flow=0,aug=INF;
    gap[0]=NV;
    while(level[vs]<NV){
        bool flag=false;
        for(int &i=cur[u]; i!=-1; i=edge[i].next){
            int v=edge[i].v;
            if(edge[i].cap!=edge[i].flow && level[u]==level[v]+1){
                flag=true;
                pre[v]=u;
                u=v;
                //aug=(aug==-1?edge[i].cap:min(aug,edge[i].cap));
                aug=min(aug,edge[i].cap-edge[i].flow);
                if(v==vt){
                    flow+=aug;
                    for(u=pre[v]; v!=vs; v=u,u=pre[u]){
                        edge[cur[u]].flow+=aug;
                        edge[cur[u]^1].flow-=aug;
                    }
                    //aug=-1;
                    aug=INF;
                }
                break;
            }
        }
        if(flag) continue;
        int minlevel=NV;
        for(int i=head[u]; i!=-1; i=edge[i].next){
            int v=edge[i].v;
            if(edge[i].cap!=edge[i].flow && level[v]<minlevel){
                minlevel=level[v];
                cur[u]=i;
            }
        }
        if(--gap[level[u]]==0) break;
        level[u]=minlevel+1;
        gap[level[u]]++;
        u=pre[u];
    }
    return flow;
}
char map1[22][22],map2[22][22];
int main(){
    int t,n,d;
    scanf("%d",&t);
    for(int cse=1; cse<=t; ++cse){
        scanf("%d%d",&n,&d);
        for(int i=0; i<n; ++i) scanf("%s",map1[i]);
        for(int i=0; i<n; ++i) scanf("%s",map2[i]);

        int tot=0,m=strlen(map1[0]);
        vs=n*m*2; vt=vs+1; NV=vt+1; NE=0;
        memset(head,-1,sizeof(head));

        for(int i=0; i<n; ++i){
            for(int j=0; j<m; ++j){
                addEdge(i*m+j,i*m+j+n*m,map1[i][j]-'0');
                for(int ni=0; ni<n; ++ni){
                    for(int nj=0; nj<m; ++nj){
                        if(ni==i && nj==j) continue;
                        if((nj-j)*(nj-j)+(ni-i)*(ni-i)<=d*d) addEdge(i*m+j+n*m,ni*m+nj,INF);
                    }
                }
                if(map2[i][j]=='L') addEdge(vs,i*m+j,1),++tot;
                if(i+1<=d||n-i<=d || j+1<=d||m-j<=d) addEdge(i*m+j+n*m,vt,INF);
            }
        }
        tot-=ISAP();
        if(tot==0) printf("Case #%d: no lizard was left behind.
",cse);
        else if(tot==1) printf("Case #%d: 1 lizard was left behind.
",cse);
        else printf("Case #%d: %d lizards were left behind.
",cse,tot);
    }
    return 0;
}