HDU 2732 最大流 拆点

 

题意：矩阵里面有些蜥蜴要跳到外面，每一个规定了可以被经过的次数，每次跳的距离小于等于d，问最后还有多少蜥蜴留下

分析：最大流，ans=总蜥蜴数-最大流

把每个格子拆分成2个点，2点之间的容量为格子能被经过的次数，然后对每个格子向它在d步内能到达的格子添加容量为INF的边，能直接跳出的格子向终点添加容量为INF的边，起点到每个有蜥蜴的格子添加容量为1的边，ISAP求最大流

#include <iostream>#include<cstdio>#include<cstring>#include<algorithm>#define INF 0x7fffffff#define M 550000#define N 1005using namespace std;int map[N][N],head[N],cnt,nv;char map1[N][N],map2[N][N];struct edge{    int v,next,cap,flow;}e[M];void init(){    memset(head,-1,sizeof(head));    cnt=0;}void addedge(int u,int v,int cap){    e[cnt].v=v;e[cnt].cap=cap;e[cnt].flow=0;e[cnt].next=head[u];head[u]=cnt++;    e[cnt].v=u;e[cnt].cap=0;e[cnt].flow=0;e[cnt].next=head[v];head[v]=cnt++;}int gap[N],dist[N],pre[N],curedge[N];int ISAP(int s,int t){    int cur_flow,u,tmp,neck,i,max_flow=0;    memset(gap,0,sizeof(gap));    memset(pre,-1,sizeof(pre));    memset(dist,0,sizeof(dist));    for(i=1;i<=nv;i++){        curedge[i]=head[i];    }    gap[nv]=nv;    u=s;    while(dist[s]<nv){        if(u==t){            cur_flow=INF;            for(i=s;i!=t;i=e[curedge[i]].v){                if(cur_flow>e[curedge[i]].cap){                    neck=i;                    cur_flow=e[curedge[i]].cap;                }            }            for(i=s;i!=t;i=e[curedge[i]].v){                tmp=curedge[i];                e[tmp].cap-=cur_flow;                e[tmp].flow+=cur_flow;                tmp^=1;                e[tmp].cap+=cur_flow;                e[tmp].flow-=cur_flow;            }            max_flow+=cur_flow;            u=neck;        }        for(i=curedge[u];i!=-1;i=e[i].next){            if(e[i].cap>0&&dist[u]==dist[e[i].v]+1)            break;        }        if(i!=-1){            curedge[u]=i;            pre[e[i].v]=u;            u=e[i].v;        }else {            if(0==--gap[dist[u]])            break;            curedge[u]=head[u];            for(tmp=nv,i=head[u];i!=-1;i=e[i].next){                if(e[i].cap>0)                tmp=tmp<dist[e[i].v]?tmp:dist[e[i].v];            }            dist[u]=tmp+1;            ++gap[dist[u]];            if(u!=s)            u=pre[u];        }    }    return max_flow;}int main(){    int total,vt,t,ics=0,n,m,sum,ans,i,j,d,x,y;    scanf("%d",&t);    while(t--){        scanf("%d%d",&n,&d);        init();        memset(map,0,sizeof(map));        for(i=0;i<n;i++){            scanf("%s",map1[i]);        }        for(i=0;i<n;i++){            scanf("%s",map2[i]);        }        m=strlen(map2[0]);        total=0;        for(i=0;i<n;i++){            for(j=0;j<m;j++){                if(map1[i][j]>'0'){//拆点                    map[i][j]=++total;                    addedge(total*2-1,total*2,map1[i][j]-'0');                }            }        }        vt=total*2+1,nv=vt+1;        sum=0;        for(i=0;i<n;i++){            for(j=0;j<m;j++){                if(map2[i][j]=='L'){                    sum++;                    addedge(0,map[i][j]*2-1,1); //源点跟有蜥蜴的点连边                }            }        }        for(i=0;i<n;i++){            for(j=0;j<m;j++){                if(map[i][j]){                    for(x=-d;x<=d;x++){                        for(y=abs(x)-d;y<=d-abs(x);y++){                            int ii=i+x;                            int jj=j+y;                            if(ii<0||jj<0||ii>=n||jj>=m)                            continue;                            if(ii==i&&jj==j)                            continue;                            if(!map[ii][jj])                            continue;                            addedge(2*map[i][j],2*map[ii][jj]-1,INF);                        }                    }                    if(i<d||j<d||i>n-1-d||j>m-1-d){                        addedge(map[i][j]*2,vt,INF);                    }                }            }        }        ans=sum-ISAP(0,vt);        printf("Case #%d: ",++ics);        if(ans)        printf("%d",ans);        else        printf("no");        if(ans>1)        printf(" lizards were left behind.\n");        else        printf(" lizard was left behind.\n");    }    return 0;}