• 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
    ",sum-res);
        return 0;
    }
    View Code
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  • 原文地址:https://www.cnblogs.com/lishiyao/p/6502431.html
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