洛谷$P$2472 蜥蜴 $[SCOI2007]$ 网络流

正解:网络流

解题报告:

传送门!

$umm$一看就是个最大流呗,,,就直接考虑怎么建图趴$QwQ$

首先看到这个高度减小其实就相当于对这个点的次数有约束,就显然拆点呗,流量为高度

然后$S$连向左侧所有有蜥蜴的点,流量为1.右侧所有边界点连向$T$,流量为$inf$.

然后就做完了?$QwQ$.

#include<bits/stdc++.h>
using namespace std;
#define il inline
#define gc getchar()
#define t(i) edge[i].to
#define n(i) edge[i].nxt
#define w(i) edge[i].wei
#define ri register int
#define rb register bool
#define rc register char
#define rp(i,x,y) for(ri i=x;i<=y;++i)
#define e(i,x) for(ri i=head[x];~i;i=n(i))

const int N=+,inf=1e9,M=;
int n,m,d,S,T,ed_cnt=-,head[N],as,cur[N],dep[N],sz,cnt,hei[M][M];
char s[N];
struct ed{int to,nxt,wei;}edge[N<<];

il int read()
{
    rc ch=gc;ri x=;rb y=;
    while(ch!='-' && (ch>'' || ch<''))ch=gc;
    if(ch=='-')ch=gc,y=;
    while(ch>='' && ch<='')x=(x<<)+(x<<)+(ch^''),ch=gc;
    return y?x:-x;
}
il int nam(ri x,ri y){return (x-)*m+y;}
il void ad(ri x,ri y,ri z){/*printf("%d -> %d : %d\n",y,x,z);*/edge[++ed_cnt]=(ed){x,head[y],z};head[y]=ed_cnt;edge[++ed_cnt]=(ed){y,head[x],};head[x]=ed_cnt;}
il bool bfs()
{
    queue<int>Q;Q.push(S);memset(dep,,sizeof(dep));dep[S]=;
    while(!Q.empty()){ri nw=Q.front();Q.pop();e(i,nw)if(w(i) && !dep[t(i)])dep[t(i)]=dep[nw]+,Q.push(t(i));}
    return dep[T];
}
il int dfs(ri nw,ri flow)
{
    if(nw==T || !flow)return flow;ri ret=;
    for(ri &i=cur[nw];~i;i=n(i))
        if(w(i) && dep[t(i)]==dep[nw]+){ri tmp=dfs(t(i),min(flow,w(i)));ret+=tmp,w(i)-=tmp;w(i^)+=tmp,flow-=tmp;}
    return ret;
}
il int dinic(){ri ret=;while(bfs()){rp(i,S,T)cur[i]=head[i];while(int d=dfs(S,inf))ret+=d;}return ret;}

int main()
{
//    freopen("2472.in","r",stdin);freopen("2472.out","w",stdout);
    memset(head,-,sizeof(head));n=read();m=read();d=read();sz=n*m;S=;T=*sz+;
    rp(i,,n){scanf("%s",s+);rp(j,,m){hei[i][j]=s[j]^'';if(!hei[i][j])continue;ri pos=nam(i,j);ad(n*m+pos,pos,hei[i][j]);}}
    rp(i,,n){scanf("%s",s+);rp(j,,m)if(s[j]=='L')ad(nam(i,j),S,),++cnt;}
    rp(i,,n)
    {
        rp(j,,m)
        {
            if(!hei[i][j])continue;
            bool gdgs=;
            rp(p,-d,d)
            {
                rp(q,abs(p)-d,d-abs(p))
                {
                    ri to_x=i+p,to_y=j+q;
                    if(to_x> && to_y> && to_x<=n && to_y<=m)ad(nam(to_x,to_y),sz+nam(i,j),inf);
                    else gdgs=;
                }
            }
            if(gdgs)ad(T,sz+nam(i,j),inf);
        }
    }
    printf("%d\n",cnt-dinic());
    return ;
}