One hundred layer

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 2172    Accepted Submission(s): 748

Problem Description

Now there is a game called the new man down 100th floor. The rules of this game is:

　　

1.  At first you are at the 1st floor. And the floor moves up. Of course you can choose which part you will stay in the first time.

　　

2.  Every floor is divided into M parts. You can only walk in a direction (left or right). And you can jump to the next floor in any part, however if you are now in part “y”, you can only jump to part “y” in the next floor! (1<=y<=M)

　　

3.  There are jags on the ceils, so you can only move at most T parts each floor.

　　

4.  Each part has a score. And the score is the sum of the parts’ score sum you passed by.

Now we want to know after you get the 100th floor, what’s the highest score you can get.

 

Input

The first line of each case has four integer N, M, X, T(1<=N<=100, 1<=M<=10000, 1<=X， T<=M). N indicates the number of layers; M indicates the number of parts. At first you are in the X-th part. You can move at most 

T parts in every floor in only one direction.

Followed N lines, each line has M integers indicating the score. (-500<=score<=500)

 

Output

Output the highest score you can get.

 

Sample Input

3 3 2 1 7 8 1 4 5 6 1 2 3

 

Sample Output

29

 

Source

题意：N*M的地图，主角初始在1,x位置，每行可以选择仅向左走<=t个位置或仅向右走<=t个位置，然后进入下一行的相应位置，求走到第N行得到的最大值。

思路：从左往右时dp[i][j] = max(dp[i][j], dp[i-1][j-k]+sum[j]-sum[j-k-1] ),k∈[0，t]，右往左时依此类推，但纯dp计算会超时，发现每次计算时sun[j]是定值，部分数据可以重复使用，因此可维护一个单调减队列优化时间。

# include 
    
     # include 
     
      # include 
      
       using namespace std;struct node{    int pos, data;};int map[103][10003], dp[103][10003], sum[10003];int main(){    int n, m, x, t, i, j, ans;    struct node tmp;    while(~scanf("%d%d%d%d",&n,&m,&x,&t))    {        ans = -0x3f3f3f3f;        list
       
        p;        for(i=1; i<=n; ++i)            for(j=1; j<=m; ++j)            {                scanf("%d",&map[i][j]);                dp[i][j] = -0x3f3f3f3f;            }        dp[1][x] = map[1][x];        for(i=x-1; i>=1&&i>=x-t; --i)            dp[1][i] = dp[1][i+1] + map[1][i];        for(i=x+1; i<=m&&i<=x+t; ++i)            dp[1][i] = dp[1][i-1] + map[1][i];        for(i=2; i<=n; ++i)        {            p.clear();            sum[0] = 0;            for(j=1; j<=m; ++j)            {                sum[j] = sum[j-1] + map[i][j];                while(!p.empty() && p.front().pos
        
         =1; --j)            {                sum[j] = sum[j+1] + map[i][j];                while(!p.empty() && p.front().pos>j+t)                    p.pop_front();                tmp.pos = j;                tmp.data = dp[i-1][j] - sum[j+1];                while(!p.empty() && p.back().data < tmp.data)                    p.pop_back();                p.push_back(tmp);                dp[i][j] = max(dp[i][j], sum[j] + p.front().data);            }        }        for(i=1; i<=m; ++i)            ans = max(ans, dp[n][i]);        printf("%d\n",ans);    }    return 0;}