洛谷 P2472 蜥蜴 （最大流）

题面

戳我

思路

有蜥蜴连超级源点，可以出图的连超级汇点，不用说了，然后就是有高度的柱子裂点，边权为高度，然后可以互相跳的连边，跑dinic就好了。

代码实现

#include<cstdio>
#include<algorithm>
#include<vector>
#include<queue>
#include<map>
#include<iostream>
#include<cstring>
#include<cmath>
using namespace std;
#define rep(i,f_start,f_end) for (int i=f_start;i<=f_end;++i)
#define per(i,n,a) for (int i=n;i>=a;i--)
#define MT(x,i) memset(x,i,sizeof(x) )
#define rev(i,start,end) for (int i=start;i<end;i++)
#define inf 0x3f3f3f3f
#define mp(x,y) make_pair(x,y)
#define lowbit(x) (x&-x)
#define exp 1e-8
#define N 1000005 
#define fi first 
#define se second
#define pb push_back
const int mod=1e9+7;
typedef long long ll;
typedef vector <int> VI;
typedef pair<int ,int> PII;
typedef pair<int ,PII> PIII;
ll gcd (ll a,ll b) {return b?gcd (b,a%b):a; }
inline int read() {
    char ch=getchar(); int x=0, f=1;
    while(ch<'0'||ch>'9') {
        if(ch=='-') f=-1;
        ch=getchar();
    } while('0'<=ch&&ch<='9') {
        x=x*10+ch-'0';
        ch=getchar();
    } return x*f;
}
const int maxn=1e4+10;
struct edge {
    int v,flow,next;
}e[500000];
int head[maxn],cnt=-1;

inline void add (int u,int v,int flow) {
    e[++cnt]=(edge) {v,flow,head[u]};
    head[u]=cnt;
}
inline void add_edge (int u,int v,int flow) {
    add (u,v,flow);
    add (v,u,0);
}

int dis[maxn],cur[maxn];
int s,t,n,m,ans,x,y;
int bfs () {
    MT (dis,0);
    queue<int > q;
    q.push (s);
    dis[s]=1;
    while (q.size ()) {
        int x=q.front (); q.pop ();
        for (int i=head[x];i!=-1;i=e[i].next) {
            if (dis[e[i].v]==0&&e[i].flow) {
                dis[e[i].v]=dis[x]+1;
                q.push (e[i].v);
            }
        }
    }
    return dis[t];
}

int dfs (int now,int nowflow) {
    if (now==t) return nowflow;
    for (int &i=cur[now];~i;i=e[i].next) {
        if (dis[e[i].v]==dis[now]+1&&e[i].flow) {
            int canflow=dfs (e[i].v,min (nowflow,e[i].flow));
            if (canflow>0) {
                e[i].flow-=canflow;
                e[i^1].flow+=canflow;
                return canflow;
            }
        }
    }
    return 0;
}

void Dinic () {
    while (bfs ()) {
       for (int i=s;i<=t;i++) cur[i]=head[i];
       while (int val=dfs(s,inf)) ans+=val;
    }
}

inline int id1 (int x,int y) {
    return (x-1)*m+y;
}

inline int id2 (int x,int y) {
    return (x-1)*m+y+n*m;
}

inline int distance (int x1,int y1,int x2,int y2) {
    return  (x1-x2)*(x1-x2)+(y1-y2)*(y1-y2);
}

int st,k,allcost;
int height[maxn][maxn];
char s1[210];
char s2[210];
int main () {
    scanf ("%d%d%d",&n,&m,&st);
    MT (head,-1);
    s=0,t=n*m*2+1;
    rep (i,1,n) {
        scanf ("%s",s1+1);
        rep (j,1,m) if (s1[j]-'0'>0) add_edge (id1 (i,j),id2 (i,j),s1[j]-'0');
    }
    rep (i,1,n) {
        scanf ("%s",s2+1);
        rep (j,1,m) if (s2[j]=='L') {
            add_edge (s,id1 (i,j),1),allcost++;
        }
    }  
    rep (i,1,n) 
      rep (j,1,m) {
          if (i-st<1||i+st>n||j-st<1||j+st>m) {
              add_edge (id2 (i,j),t,inf);
          }
      }
    rep (x1,1,n) 
     rep (y1,1,m) 
      rep (x2,1,n)
       rep (y2,1,m) {
           if (distance (x1,y1,x2,y2)<=st*st) {
               add_edge (id2 (x1,y1),id1 (x2,y2),inf);
           }
       }
    Dinic ();
    printf ("%d
",allcost-ans);
    return 0;
}