【Luogu】P2598狼和羊的故事(最小割转最大流)

  题目链接

  最小割水题。入点向白点连边,白点向白点、黑点和空点连边,空点向空点和黑点连边,黑点向黑点和汇点连边。然后跑最大流即可。

  话说Fd最近怎么光做水题啊……一点用都没有……qwq

  我太菜了,做完一道题就打十来分钟osu……这么颓下去吃枣药丸

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cctype>
#include<cstdlib>
#include<queue>
#define maxn 10020
#define maxm 200200
using namespace std;

int u[5]={0,1,0,-1,0};
int w[5]={0,0,1,0,-1};
int n,m;

inline long long read(){
    long long num=0,f=1;
    char ch=getchar();
    while(!isdigit(ch)){
        if(ch=='-')    f=-1;
        ch=getchar();
    }
    while(isdigit(ch)){
        num=num*10+ch-'0';
        ch=getchar();
    }
    return num*f;
}

struct Edge{
    int next,to,val;
}edge[maxm];
int head[maxn],num;
inline void addedge(int from,int to,int val){
    edge[++num]=(Edge){head[from],to,val};
    head[from]=num;
}
inline void add(int from,int to,int val){
    addedge(from,to,val);
    addedge(to,from,0);
}

inline int count(int i){    return i&1?i+1:i-1;    }
inline int calc(int i,int j){    return (i-1)*m+j;    }

bool vis[maxn];
int dfn[maxn];
int list[maxn];
int Start,End;

bool bfs(){
    memset(vis,0,sizeof(vis));
    queue<int>q;    q.push(Start);    vis[Start]=1;    dfn[Start]=1;
    while(!q.empty()){
        int from=q.front();    q.pop();
        for(int i=head[from];i;i=edge[i].next){
            int to=edge[i].to;
            if(vis[to]||edge[i].val==0)    continue;
            vis[to]=1;    dfn[to]=dfn[from]+1;
            q.push(to);
        }
    }
    return vis[End];
}

int dfs(int x,int val){
    if(x==End||val==0)    return val;
    int flow=0;    vis[x]=1;
    for(int &i=list[x];i;i=edge[i].next){
        int to=edge[i].to;
        if(vis[to]||edge[i].val==0||dfn[to]!=dfn[x]+1)    continue;
        int now=dfs(to,min(val,edge[i].val));
        val-=now;    edge[i].val-=now;    edge[count(i)].val+=now;    flow+=now;
        if(val<=0)    break;
    }
    if(flow!=val)    dfn[x]=-1;
    return flow;
}

inline int maxflow(){
    int ans=0;
    while(bfs()){
        memset(vis,0,sizeof(vis));
        for(int i=Start;i<=End;++i)    list[i]=head[i];
        int now=dfs(Start,0x7fffffff);
        if(now==0)    break;
        ans+=now;
    }
    return ans;
}

int q[202][202];

int main(){
    n=read(),m=read();    End=n*m+1;
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j){
            q[i][j]=read();
            if(q[i][j]!=2)    q[i][j]=1-q[i][j];
            if(q[i][j]==2)        add(Start,calc(i,j),0x7fffffff);
            else if(q[i][j]==0)    add(calc(i,j),End,0x7fffffff);
        }
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j){
            int now=calc(i,j);
            for(int k=1;k<=4;++k){
                int a=i+u[k],b=j+w[k];
                if(a&&b&&a<=n&&b<=m){
                    int ret=calc(a,b);
                    if(q[i][j]>=q[a][b])    add(now,ret,1);
                }
            }
        }
    printf("%d",maxflow());
    return 0;
}

 

posted @ 2018-01-23 15:00  Konoset  阅读(117)  评论(0编辑  收藏  举报