752. [BJOI2006] 狼抓兔子

★★★☆   输入文件:bjrabbit.in   输出文件:bjrabbit.out   简单对比

时间限制:1 s   内存限制:162 MB

Description   Source: Beijing2006 [BJOI2006]

八中OJ上本题链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1001

现在小朋友们最喜欢的"喜羊羊与灰太狼",话说灰太狼抓羊不到,但抓兔子还是比较在行的,而且现在的兔子还比较笨,它们只有两个窝,现在你做为狼王,面对下面这样一个网格的地形:

 

左上角点为(1,1),右下角点为(N,M)(上图中N=4,M=5).有以下三种类型的道路 1:(x,y)<==>(x+1,y) 2:(x,y)<==>(x,y+1) 3:(x,y)<==>(x+1,y+1) 道路上的权值表示这条路上最多能够通过的兔子数,道路是无向的. 左上角和右下角为兔子的两个窝,开始时所有的兔子都聚集在左上角(1,1)的窝里,现在它们要跑到右下解(N,M)的窝中去,狼王开始伏击这些兔子.当然为了保险起见,如果一条道路上最多通过的兔子数为K,狼王需要安排同样数量的K只狼,才能完全封锁这条道路,你需要帮助狼王安排一个伏击方案,使得在将兔子一网打尽的前提下,参与的狼的数量要最小。因为狼还要去找喜羊羊麻烦.

Input

第一行为N,M.表示网格的大小,N,M均小于等于1000.接下来分三部分 第一部分共N行,每行M-1个数,表示横向道路的权值. 第二部分共N-1行,每行M个数,表示纵向道路的权值. 第三部分共N-1行,每行M-1个数,表示斜向道路的权值. 输入文件保证不超过10M

Output

输出一个整数,表示参与伏击的狼的最小数量.

Sample Input

3 4
5 6 4
4 3 1
7 5 3
5 6 7 8
8 7 6 5
5 5 5
6 6 6

Sample Output

14

 裸地网络流没调出来

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#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>
 
using namespace std;
const int N = 1e6 + 10;
const int Maxn = 99999999;
 
int head[N], dis[N];
int n, m, S, T, now, w, x;
struct Node{
    int u, v, cap, flow, nxt;
}E[N<<1-1];
queue <int> Q;
 
inline int read()
{
    int x = 0, f = 1;
    char c = getchar();
    while(c < '0' || c > '9')
    {
        if(c == '-')
            f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9')
    {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
 
inline void add(int u, int v, int cap)
{
    E[now].v = v;
    E[now].cap = cap;
    E[now].flow = 0;
    E[now].nxt = head[u];
    head[u] = now++;
}
 
inline bool bfs()
{
    for(int i = 1; i <= T; i ++)
        dis[i] = -1;
    dis[S] = 0;
    Q.push(S);
    while(!Q.empty())
    {
        int topp = Q.front();
        Q.pop();
        for(int i = head[topp]; ~ i; i = E[i].nxt)
        {
            if(dis[E[i].v] == -1 && E[i].cap - E[i].flow > 0)
            {
                dis[E[i].v] = dis[topp] + 1;
                Q.push(E[i].v);
            }
        }
    }
    if(dis[T] == -1)
        return 0;
    else
        return 1;
}
 
int dfs(int start,int minn)
{
    if(start == T /*|| minn <= 0*/)
        return minn;
    int ret = 0, flo;
    for(int i = head[start]; ~ i; i = E[i].nxt)
    {
        if(dis[E[i].v] == dis[start] + 1 && E[i].cap - E[i].flow > 0)
        {
            flo = dfs(E[i].v, min(minn, E[i].cap - E[i].flow));
            E[i].flow += flo;
            E[i ^ 1].flow -= flo;
            ret += flo;
            minn -= flo;
        }
    }
    return ret;
}
 
inline void Dinic()
{
    int answer = 0;
    while(bfs())
        answer += dfs(S, Maxn);
    printf("%d",answer);
}
 
int main()
{
    //freopen("bjrabbit.in","r",stdin);
    //freopen("bjrabbit.out","w",stdout);
    n = read();
    m = read();
    S = 1;
    T = n * m;
    for(int i = 1; i <= T; i ++)
        head[i] = -1;
    for(int i = 1; i <= n; i ++)
        for(int j = 1; j <= m - 1; j ++)
        {
            w = read();
            x = (i - 1) * n + j + i - 1;
            add(x, x + 1, w),
            add(x + 1, x, 0);
            //cout<<x<<" "<<x+1<<endl;
        }
    for(int i = 1; i <= n - 1; i ++)
        for(int j = 1; j <= m ; j ++)
        {
            w = read();
            x = (i - 1) * n + j + i - 1;
            add(x, x + m, w);
            add(x + m, x, 0);
            //cout<<x<<" "<<x+m<<endl;
        }
         
    for(int i = 1; i <= n - 1; i ++)
        for(int j = 1; j <= m - 1; j ++)
        {
            w = read();
            x = (i - 1) * n + j + i - 1;
            add(x, x + m + 1, w);
            add(x + m + 1, x, 0);
            //cout<<x<<" "<<x+m+1<<endl;
        }  
    Dinic();
    return 0;
}
/* 
3 4
5 6 4
4 3 1
7 5 3
5 6 7 8
8 7 6 5
5 5 5
6 6 6
*/

对偶图最短路

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//平面图的最小割 = 对偶图的最短路
 
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 2001000
#define inf 0x7fffffff
 
struct Edge{
    int v,next,k;
}edge[N<<2];
int n,m,ans,num=0,S,T,head[N],dis[N],q[N*5];
bool vis[N];
 
int in()
{
    int x=0; char ch=getchar();
    while (ch<'0' || ch>'9') ch=getchar();
    while (ch>='0' && ch<='9') x=x*10+ch-'0',ch=getchar();
    return x;
}
 
void add(int u,int v,int k)
{
    edge[++num].v=v; edge[num].k=k;
    edge[num].next=head[u]; head[u]=num;
}
 
void spfa()
{
    int h=0,t=1;
    memset(dis,127/3,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dis[S]=0,vis[S]=1,q[h]=S;
    while (h<t)
    {
        int u=q[h]; vis[u]=0; h++;
        if (h>=N*5) h=0,t=1;
        for (int i=head[u]; i; i=edge[i].next)
        {
            int v=edge[i].v;
            if (dis[v]>dis[u]+edge[i].k)
            {
                dis[v]=dis[u]+edge[i].k;
                if (!vis[v]) vis[v]=1,q[t++]=v;
            }
        }
    }
    ans=dis[T];
}
 
void build()
{
    S=0,T=((n-1)*(m-1))<<1|1;
    for (int i=1; i<=n; i++)
        for (int j=1; j<m; j++)
        {
            int x=in(),u,v;
            if (i==1) u=S,v=j;
            else if (i==n) u=((i-2)<<1|1)*(m-1)+j,v=T;
            else u=((i-2)<<1|1)*(m-1)+j,v=((i-1)<<1)*(m-1)+j;
            add(u,v,x),add(v,u,x);
        }
    for (int i=1; i<n; i++)
        for (int j=1; j<=m; j++)
        {
            int x=in(),u,v;
            if (j==1) u=((i-1)<<1|1)*(m-1)+1,v=T;
            else if (j==m) u=S,v=((i-1)<<1|1)*(m-1);
            else u=((i-1)<<1)*(m-1)+j-1,v=((i-1)<<1|1)*(m-1)+j;
            add(u,v,x),add(v,u,x);
        }
    for (int i=1; i<n; i++)
        for (int j=1; j<m; j++)
        {
            int x=in(),u,v;
            u=((i-1)<<1)*(m-1)+j,v=((i-1)<<1|1)*(m-1)+j;
            add(u,v,x),add(v,u,x);
        }
}
 
int main()
{
    freopen("bjrabbit.in","r",stdin);
    freopen("bjrabbit.out","w",stdout);
    n=in(),m=in();
 
    if (n==1 || m==1)
    {
        if (n>m) swap(n,m);
        ans=inf;
        for (int i=1; i<m; i++)
        {
            int x=in();
            ans=min(ans,x);
        }
        if (ans==inf) ans=0;
        printf("%d\n",ans);
        return 0;
    }
 
    build(); spfa();
    printf("%d\n",ans);
    return 0;
}

  

posted @   ioioioioioio  阅读(149)  评论(0编辑  收藏  举报
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