HDU 4374 One hundred layer DP的单调队列优化

Posted 2020-07-12 Meek

One hundred layer

 

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

 

题意：

 

　　给你n*m的图，起始位置在第一行的第x个位置，现在你可以选择一个方向至多走T个位置然后走向下一行，直到第n行

　　路过的格子里的值总和最大是多少

题解：

　　首先想到dp[i][j]表示到达当前(i,j)格子的最大答案，那么最后答案显然了

　　思考如何得到(i,j)的最优答案

　　他可以是从左边上一层走下来再向右走到j位置，且走过不超过T，也可以是右边

　　对于左边：dp[i][j] = dp[i-1][k] - sum[i][k-1] + sum[i][j];

　　dp[i-1][k] - sum[i][k-1]这一块是上一层，和当前层没有任何关系，我们可以预处理啊

　　那么我们维护一个大小T的单调队列，左边右边扫一波就好了

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <cmath>
#include<queue>
using namespace std;
const int N = 120, M = 1e4+20, mod = 1000000007,inf = 2e9;
typedef long long ll;

int n,m,x,t,a[N][M],sum[N][M],dp[N][M];
int main() {

    while(scanf("%d%d%d%d",&n,&m,&x,&t)!=EOF) {
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++) scanf("%d",&a[i][j]);
        for(int i=1;i<=n;i++) {
            sum[i][0] = 0;
            for(int j=1;j<=m;j++) sum[i][j] = sum[i][j-1] + a[i][j];
        }
        for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) dp[i][j] = -inf;
        for(int i=x;i>=1&&i>=x-t;i--)
            dp[1][i] = sum[1][x] - sum[1][i-1];
        for(int i=x;i<=m&&i<=x+t;i++)
            dp[1][i] = sum[1][i] - sum[1][x-1];

        deque<int >q;
        for(int i=2;i<=n;i++) {
            //从左向右

            while(!q.empty()) q.pop_back();
            dp[i][1] = dp[i-1][1] + a[i][1];
            q.push_back(1);
            for(int j=2;j<=m;j++) {
                while(!q.empty()&&j-q.front()>t) q.pop_front();
                int now = dp[i-1][j] - sum[i][j-1];
                while(!q.empty()&&dp[i-1][q.back()]-sum[i][q.back()-1]<=now) q.pop_back();
                q.push_back(j);
                int  pos = q.front();
                dp[i][j] = max(dp[i][j],dp[i-1][pos]-sum[i][pos-1]+sum[i][j]);
            }

            while(!q.empty()) q.pop_back();
            q.push_back(m);
            dp[i][m] = max(dp[i][m],dp[i-1][m]+a[i][m]);

            for(int j=m-1;j>=1;j--) {
                while(!q.empty()&&q.front()-j>t) q.pop_front();
                int now = dp[i-1][j] + sum[i][j];
                while(!q.empty()&&dp[i-1][q.back()]+sum[i][q.back()]<=now) q.pop_back();
                q.push_back(j);
                int pos = q.front();
                dp[i][j] = max(dp[i][j],dp[i-1][pos]+sum[i][pos]-sum[i][j-1]);
            }

        }
        int ans = -inf;
        for(int i=1;i<=m;i++) ans = max(ans,dp[n][i]);
        printf("%d\n",ans);
    }
    return 0;
}