【bzoj1066】: [SCOI2007]蜥蜴 图论-最大流

【bzoj1066】: [SCOI2007]蜥蜴

把石柱拆点，流量为高度

然后S与蜥蜴连流量1的边

互相能跳到的石柱连inf的边

石柱能到边界外的和T连inf的边

然后跑dinic就好了

/* http://www.cnblogs.com/karl07/ */
#include <cstdlib>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
using namespace std;
const int N=1000000;
const int inf=1e9;
struct edge{
    int next,to,c;
}e[N+1];
int ade=1,S=1,T=2,n,m,c,d,CNT=2;
int first[N+1],dis[N+1],now[N+1],mp[25][25],I[25][25],O[25][25];
queue <int> Q;

void addedge(int x,int y,int cap){
    e[++ade].next=first[x];
    e[ade].to=y;
    e[ade].c=cap;
    first[x]=ade;
    e[++ade].next=first[y];
    e[ade].to=x;
    e[ade].c=0;
    first[y]=ade;
}

#define s e[x].to
#define cap e[x].c
#define CAP e[x^1].c
bool bfs(){
    for (int i=1;i<=CNT;i++) dis[i]=inf;
    Q.push(S);
    dis[S]=0;
    while (!Q.empty()){
        int p=Q.front();
        Q.pop();
        for (int x=first[p];x;x=e[x].next){
            if (dis[s]>dis[p]+1 && cap>0){
                Q.push(s);
                dis[s]=dis[p]+1;
            }
        }
    }
    return dis[T]!=inf;
}

int dfs(int p,int mf){
    if (p==T) return mf;
    for (int x=now[p];x;x=e[x].next){
        now[p]=x;
        if (dis[s]==dis[p]+1 && cap>0){
            int f=dfs(s,min(mf,cap));
            if (f){
                cap-=f;
                CAP+=f;
                return f;
            }
        }
    }
    return 0;
}
#undef s
#undef cap
#undef CAP


int dinic(){
    int ans=0,f;
    while (bfs()){
        for (int i=1;i<=CNT;i++) now[i]=first[i];
        while (f=dfs(S,inf)){
            ans+=f;
        }
    }
    return ans;
}

void make_p(){
    for (int i=1;i<=n;i++){
        for (int j=1;j<=m;j++){
            if (mp[i][j]){
                addedge(I[i][j],O[i][j],mp[i][j]);
                for (int x=i-d;x<=i+d;x++){
                    for (int y=j-d;y<=j+d;y++){
                        if (abs(i-x)+abs(j-y)>d) continue;
                        if (x<1 || x>n || y<1 || y>m) {addedge(O[i][j],T,inf); continue;}
                        if (mp[x][y] && (x!=i || y!=j)){addedge (O[i][j],I[x][y],inf); }
                    }
                }
            }
        }
    }
}

int main(){
    scanf("%d%d%d",&n,&m,&d);
    for (int i=1;i<=n;i++){
        char s[25];
        scanf("%s",s+1);
        for (int j=1;j<=m;j++){
            mp[i][j]=s[j]-'0';
            if (mp[i][j]) I[i][j]=++CNT,O[i][j]=++CNT;
        }
    }
    make_p();
    for (int i=1;i<=n;i++){
        char s[25];
        scanf("%s",s+1);
        for (int j=1;j<=m;j++){
            if (s[j]=='L'){
                addedge(S,I[i][j],1);
                c++;
            }
        }
    }
    printf("%d
",c-dinic());
    return 0;
}