BZOJ 1066 蜥蜴(网络流)

很普通的拆点网络流,把每个柱子拆成两个点(i,j,0)和(i,j,1).对于柱子的高度限制则加边((i,j,0),(i,j,1),height).

两个柱子能互相到达则加边((i,j,1),(i1,j1,0),INF). 能到达边界的柱子加边((i,j,1),t,INF).有蜥蜴的柱子加边(s,(i,j,0),1).

跑一遍最大流,答案就是总蜥蜴数-最大流。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-9
# define MOD 12345678
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int res=0, flag=0;
    char ch;
    if((ch=getchar())=='-') flag=1;
    else if(ch>='0'&&ch<='9') res=ch-'0';
    while((ch=getchar())>='0'&&ch<='9')  res=res*10+(ch-'0');
    return flag?-res:res;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=2005;
//Code begin...
 
struct Edge{int p, next, w;}edge[650005];
int head[805], cnt=2, s, t, vis[805];
char s1[25][25], s2[25][25];
queue<int>Q;
 
void add_edge(int u, int v, int w)
{
    edge[cnt].p=v; edge[cnt].next=head[u]; edge[cnt].w=w; head[u]=cnt++;
    edge[cnt].p=u; edge[cnt].next=head[v]; edge[cnt].w=0; head[v]=cnt++;
}
int bfs(void)
{
    int i, v;
    mem(vis,-1);
    while (!Q.empty()) Q.pop();
    vis[s]=0; Q.push(s);
    while (!Q.empty()) {
        v=Q.front(); Q.pop();
        for (i=head[v]; i; i=edge[i].next) {
            if (edge[i].w>0 && vis[edge[i].p]==-1) {
                vis[edge[i].p]=vis[v] + 1;
                Q.push(edge[i].p);
            }
        }
    }
    return vis[t]!=-1;
}
int dfs(int x, int low)
{
    int i, a, temp=low;
    if (x==t) return low;
    for (i=head[x]; i; i=edge[i].next) {
        if (edge[i].w>0 && vis[edge[i].p] == vis[x] + 1){
            a=dfs(edge[i].p,min(edge[i].w,temp));
            temp-=a; edge[i].w-=a; edge[i^1].w += a;
            if (temp==0) break;
        }
    }
    if (temp==low) vis[x]=-1;
    return low-temp;
}
int main ()
{
    int n, m, d, sum=0;
    scanf("%d%d%d",&n,&m,&d);
    FO(i,0,n) scanf("%s",s1[i]+1);
    FO(i,0,n) scanf("%s",s2[i]+1);
    s=0, t=2*n*m+1;
    FO(i,0,n) FOR(j,1,m) {
        if (s1[i][j]!='0') add_edge(i*m+j,i*m+j+n*m,s1[i][j]-'0');
        if (s2[i][j]=='L') add_edge(s,i*m+j,1), ++sum;
        if (i<d||(n-1-i)<d||j<=d||(m-j+1)<=d) add_edge(i*m+j+n*m,t,INF);
        FO(k,0,n) FOR(l,1,m) {
            if (k==i&&l==j) continue;
            if (abs(k-i)+abs(l-j)<=d) add_edge(i*m+j+n*m,k*m+l,INF);
        }
    }
    int res=0, temp;
    while (bfs()) while (temp=dfs(s,INF)) res+=temp;
    printf("%d\n",sum-res);
    return 0;
}
View Code

 

posted @ 2017-03-04 19:38  free-loop  阅读(100)  评论(0编辑  收藏