HDU 3657 Game (SAP）

Game

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 691    Accepted Submission(s): 277

Problem Description

onmylove has invented a game on n × m grids. There is one positive integer on each grid. Now you can take the numbers from the grids to make your final score as high as possible. The way to get score is like
the following:
● At the beginning, the score is 0;
● If you take a number which equals to x, the score increase x;
● If there appears two neighboring empty grids after you taken the number, then the score should be decreased by 2(x&y). Here x and y are the values used to existed on these two grids. Please pay attention that "neighboring grids" means there exits and only exits one common border between these two grids.

Since onmylove thinks this problem is too easy, he adds one more rule:
● Before you start the game, you are given some positions and the numbers on these positions must be taken away.
Can you help onmylove to calculate: what's the highest score onmylove can get in the game?

 

 

Input

Multiple input cases. For each case, there are three integers n, m, k in a line.
n and m describing the size of the grids is n ×m. k means there are k positions of which you must take their numbers. Then following n lines, each contains m numbers, representing the numbers on the n×m grids.Then k lines follow. Each line contains two integers, representing the row and column of one position
and you must take the number on this position. Also, the rows and columns are counted start from 1.
Limits: 1 ≤ n, m ≤ 50, 0 ≤ k ≤ n × m, the integer in every gird is not more than 1000.

 

 

Output

For each test case, output the highest score on one line.

 

 

Sample Input

2 2 1 2 2 2 2 1 1 2 2 1 2 7 4 1 1 1

 

 

Sample Output

4 9

Hint

As to the second case in Sample Input, onmylove gan get the highest score when calulating like this: 2 + 7 + 4 － 2 × (2&4) － 2 × (2&7) = 13 － 2 × 0 － 2 × 2 = 9.

 

 

Author

onmylove

 

 

Source

2010 Asia Regional Chengdu Site —— Online Contest

 

 

Recommend

lcy

 

 

 

题意: 给定一个 n * m 大小的矩阵, 矩阵中每个格子都有一个不大于 1000 的正整数,
 *       现在要从矩阵中选出若干个格子, 使得得分最大. 得分等于选取的格子里的数字
 *       和. 如果有两个格子相邻, 则得分将减去 2 * (x & y), x 和 y 为相邻两个格
 *       子内的数字. 此外, 还有一些格子是必选的.
 * 解法: 此题是个最大流算法, 和 NOI 国家集训队论文上的最大流题目类似. 建图方法
 *       是设置一个超级源点和一个超级汇点, 将横坐标和纵坐标之和为偶数的格子和源
 *       点相连, 边权值为格子的数值. 将横坐标和纵坐标之和为奇数的格子和汇点相连,
 *       边权值为格子的数值, 如果两个格子相邻, 则在它们所代表的顶点之间连一条边,
 *       权值为 2 * (x & y) . 对于必选的顶点, 将它们和源点或者汇点相连的边权值
 *       改为正无穷, 最大得分数为所有格子数值相加减去图的最大流

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int VM=101000;   //所谓的TLE都是被这个数据开小了导致的，，，God
const int EM=500100;
const int INF=0x3f3f3f3f;

struct Edge{
    int to,nxt;
    int cap;
}edge[EM<<1];

int n,m,k,cnt,head[VM],map[110][110];
int dep[VM],gap[VM],cur[VM],aug[VM],pre[VM];

void addedge(int cu,int cv,int cw){
    edge[cnt].to=cv;  edge[cnt].cap=cw;  edge[cnt].nxt=head[cu];
    head[cu]=cnt++;
    edge[cnt].to=cu;  edge[cnt].cap=0;   edge[cnt].nxt=head[cv];
    head[cv]=cnt++;
}

int src,des;

int SAP(int n){
    int max_flow=0,u=src,v;
    int id,mindep;
    aug[src]=INF;
    pre[src]=-1;
    memset(dep,0,sizeof(dep));
    memset(gap,0,sizeof(gap));
    gap[0]=n;
    for(int i=0;i<=n;i++)
        cur[i]=head[i]; // 初始化当前弧为第一条弧
    while(dep[src]<n){
        int flag=0;
        if(u==des){
            max_flow+=aug[des];
            for(v=pre[des];v!=-1;v=pre[v]){     // 路径回溯更新残留网络
                id=cur[v];
                edge[id].cap-=aug[des];
                edge[id^1].cap+=aug[des];
                aug[v]-=aug[des];   // 修改可增广量，以后会用到
                if(edge[id].cap==0) // 不回退到源点，仅回退到容量为0的弧的弧尾
                    u=v;
            }
        }
        for(int i=cur[u];i!=-1;i=edge[i].nxt){
            v=edge[i].to;    // 从当前弧开始查找允许弧
            if(edge[i].cap>0 && dep[u]==dep[v]+1){  // 找到允许弧
                flag=1;
                pre[v]=u;
                cur[u]=i;
                aug[v]=min(aug[u],edge[i].cap);
                u=v;
                break;
            }
        }
        if(!flag){
            if(--gap[dep[u]]==0)    /* gap优化，层次树出现断层则结束算法 */
                break;
            mindep=n;
            cur[u]=head[u];
            for(int i=head[u];i!=-1;i=edge[i].nxt){
                v=edge[i].to;
                if(edge[i].cap>0 && dep[v]<mindep){
                    mindep=dep[v];
                    cur[u]=i;   // 修改标号的同时修改当前弧
                }
            }
            dep[u]=mindep+1;
            gap[dep[u]]++;
            if(u!=src)  // 回溯继续寻找允许弧
                u=pre[u];
        }
    }
    return max_flow;
}

int main(){

    //freopen("input.txt","r",stdin);

    while(~scanf("%d%d%d",&n,&m,&k)){
        cnt=0;
        memset(head,-1,sizeof(head));
        src=0; des=n*m+1;
        int sum=0;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++){
                scanf("%d",&map[i][j]);
                sum+=map[i][j];
            }
        int x,y;
        while(k--){
            scanf("%d%d",&x,&y);
            if((x+y)%2==0)
                addedge(src,(x-1)*m+y,INF);
            else
                addedge((x-1)*m+y,des,INF);
        }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++){
                int tmp=(i-1)*m+j;
                if((i+j)%2==0)
                    addedge(src,tmp,map[i][j]);
                else
                    addedge(tmp,des,map[i][j]);
            }
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                if((i+j)%2==0){
                    int tmp=(i-1)*m+j;
                    if(i>1)    addedge(tmp,tmp-m,2*(map[i][j]&map[i-1][j]));
                    if(i<n)    addedge(tmp,tmp+m,2*(map[i][j]&map[i+1][j]));
                    if(j>1)    addedge(tmp,tmp-1,2*(map[i][j]&map[i][j-1]));
                    if(j<m)    addedge(tmp,tmp+1,2*(map[i][j]&map[i][j+1]));
                }
        printf("%d\n",sum-SAP(des+1));
    }
    return 0;
}