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

3 375 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;}