HDU：1569：方格取数(2)（最小割）

方格取数(2)

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6708    Accepted Submission(s): 2155

Problem Description

给你一个m*n的格子的棋盘，每个格子里面有一个非负数。
从中取出若干个数，使得任意的两个数所在的格子没有公共边，就是说所取数所在的2个格子不能相邻，并且取出的数的和最大。

 

Input

包括多个测试实例，每个测试实例包括2整数m,n和m*n个非负数(m<=50,n<=50)

 

Output

对于每个测试实例，输出可能取得的最大的和

 

Sample Input

  

   3 3
75 15 21 
75 15 28 
34 70 5
  

 

Sample Output

  

   188
  

 

Author

ailyanlu

 

Source

Happy 2007

思路：神奇的最小割，将格子分成黑白两种色，相邻两个格子颜色不一样，那么开始建图：

超级源点-黑点，权值为点权①

黑点到相邻的白点，权值INF②

所有白点到超级汇点，权值为点权③

那么求最小割就肯定是从①和③中砍掉一些边，使得源点到汇点断流，那么跑一遍最大流即可。

# include <iostream>
# include <cstring>
# include <cstdio>
# include <algorithm>
using namespace std;
const int MAXN = 2e4;
const int MAXM = 4e5+30;
const int INF = 0x3f3f3f3f;
inline void s(int &ret)
{
    char c; ret=0;
    while((c=getchar())<'0'||c>'9');
    while(c>='0'&&c<='9') ret=ret*10+(c-'0'),c=getchar();
}
inline void out(int x)
{
    if(x>9) out(x/10);
    putchar(x%10+'0');
}
struct Edge
{
    int to, next;
    int cap, flow;
}edge[MAXM];
int tot, head[MAXN], gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];
void init()
{
    tot = 0;
    memset(head, -1, sizeof(head));
}
void add(int u, int v, int w, int rw=0)
{
    edge[tot] = {v, head[u], w, 0};
    head[u] = tot++;
    edge[tot] = {u, head[v], rw, 0};
    head[v] = tot++;
}
int SAP(int start, int End, int N)
{
    memset(gap, 0, sizeof(gap));
    memset(dep, 0, sizeof(dep));
    memcpy(cur, head, sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == End)
        {
            int Min = INF;
            for(int i = pre[u];i != -1; i = pre[edge[i^1].to])
            if(Min > edge[i].cap - edge[i].flow)
                Min = edge[i].cap - edge[i].flow;
            for(int i = pre[u];i != -1; i = pre[edge[i^1].to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        bool flag = false;
        int v;
        for(int i = cur[u]; i != -1;i = edge[i].next)
        {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v]+1 == dep[u])
            {
                flag = true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if(flag)
        {
            u = v;
            continue;
        }
        int Min = N;
        for(int i = head[u]; i != -1;i = edge[i].next)
        if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
        {
            Min = dep[edge[i].to];
            cur[u] = i;
        }
        gap[dep[u]]--;
        if(!gap[dep[u]]) return ans;
        dep[u] = Min+1;
        gap[dep[u]]++;
        if(u != start) u = edge[pre[u]^1].to;
    }
    return ans;
}
int main()
{
    int m, n, k;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        int sum = 0;
        for(int i=1; i<=n; ++i)
        {
            for(int j=1; j<=m; ++j)
            {
                s(k);
                sum += k;
                if((i+j)&1)
                {
                    add(0, i*m+j, k);
                    if(i>1) add(i*m+j, (i-1)*m+j, INF);
                    if(i<n) add(i*m+j, (i+1)*m+j, INF);
                    if(j>1) add(i*m+j, i*m+j-1, INF);
                    if(j<m) add(i*m+j, i*m+j+1, INF);
                }
                else
                    add(i*m+j, (n+1)*(m+1), k);
            }
        }
        out(sum-SAP(0, (n+1)*(m+1), n*m+2));
        puts("");
    }
    return 0;
}