BZOJ 1001 狼抓兔子(网络流)

题解:这个建图很简单,只要把(1,1)这个点作为超级源,(n,m)作为超级源就可以xjbp。空间要算好。dinic当前弧优化一下就可以跑1500ms

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
#include <map>
#include <queue>
#include <vector>
#include <cstring>
#include <iomanip>
#include <set>
#include<ctime>
//#include<unordered_map>
//CLOCKS_PER_SEC
#define se second
#define fi first
#define ll long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define Pii pair<int,int>
#define Pli pair<ll,int>
#define ull unsigned long long
#define pb push_back
#define fio ios::sync_with_stdio(false);cin.tie(0)
const int N=6e6+10;
const ull base=163;
const int INF=0x3f3f3f3f;
using namespace std;
int n,m,s,t;
int head[1000010],to[N],nx[N],cur[1000010];
int cap[N];
int tot=0;
int d[1000010];
inline int read(){
    int x=0;char ch=getchar();
    while (ch<'0'||ch>'9') ch=getchar();
    while (ch<='9'&&ch>='0'){x=x*10+ch-'0';ch=getchar();}
    return x;
}
void add(int u,int v,int c){
    to[tot]=v;
    cap[tot]=c;
    nx[tot]=head[u];
    head[u]=tot++;
    
    to[tot]=u;
    cap[tot]=c;
    nx[tot]=head[v];
    head[v]=tot++;
}
void init(){
    tot=0;
    memset(head,-1,sizeof(head));
}

int bfs(){
    memset(d,-1,sizeof(d));
    queue<int>q;
    q.push(s);
    d[s]=1;
    while(!q.empty()){
        int u=q.front();q.pop();
        for(int i=head[u];~i;i=nx[i]){
            int v=to[i];
            if(d[v]==-1&&cap[i]>0){
                d[v]=d[u]+1;
                q.push(v);
            }
        }
    }
    return d[t]!=-1;
}
int dfs(int s,int a){
    if(s==t||a==0)return a;
    int flow=0,f;
    for(int &i=cur[s];~i;i=nx[i]){
        int v=to[i];
        if(d[s]+1==d[v] && cap[i]>0 && (f=dfs(v,min(a,cap[i])))>0){
            flow+=f;
            cap[i]-=f;
            cap[i^1]+=f;
            a-=f;
            if(a==0)break;
        }
    }
    return flow;
}
int dinic(){
    int ans=0;
    while(bfs()){
        for(int i=0;i<=t;i++)cur[i]=head[i];
        while(int di=dfs(s,INF)){
            ans+=di;
        }
    }
    return ans;
}
int getnum(int i,int j){
    return (i-1)*m+j;
}
int main(){
    n=read(),m=read();
    s=1,t=n*m;
    memset(head,-1,sizeof(head));
    int x,l1,l2;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=m-1;j++){
            x=read();
            l1=getnum(i,j),l2=l1+1;
            add(l1,l2,x);
        }
    }
    for(int i=1;i<=n-1;i++){
        for(int j=1;j<=m;j++){
            x=read();
            l1=getnum(i,j),l2=l1+m;
            add(l1,l2,x);
        }
    }
    for(int i=1;i<=n-1;i++){
        for(int j=1;j<=m-1;j++){
            x=read();
            l1=getnum(i,j),l2=l1+m+1;
            add(l1,l2,x);
        }
    }
    cout<<dinic()<<endl;
    return 0;
}

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转载自www.cnblogs.com/Mrleon/p/9052790.html
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