Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2494 Accepted Submission(s): 878
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
解题思路
明显可以看出dp式
dp[i][j] = dp[i - 1][k] + sum[i][j] - sum[i][k - 1];
dp[i][j] = dp[i - 1][k] + sum[i][k] - sum[i][j - 1];
但是由于 k太大了直接做会超时，由于sum[i][j]和sum[i][j - 1]都是常数， 所以利用单调队列来维护
dp[i - 1][k] - sum[i][k - 1] 和 dp[i - 1][k] + sum[i][k]
#define _CRT_SECURE_NO_WARNINGS
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <string>
#include <map>
#include <stack>
#define M 1000000
#define ll long long
#define se second
#define fi first
#define pb push_back
#define INF 0x3f3f3f3f
#define de(x) cout << #x << " = "<< x << endl;
#define eps 1e-6
using namespace std;
int n, m, x, t;
int dp[105][10005], a[105][10005], sum[105][10005];
deque <pair<int, int > > q;
int main()
{
while (cin >> n >> m >> x >> t)
{
memset(sum, 0, sizeof(sum));
memset(dp, 0, sizeof(dp));
memset(a, 0, sizeof(a));
for (int i = 1; i <= n; i++)
{
for (int j = 1; j <= m; j++)
{
scanf("%d", &a[i][j]);
sum[i][j] = sum[i][j - 1] + a[i][j];
}
}
for (int i = 0; i <= n; i++)
for (int j = 0; j <= m; j++) dp[i][j] = -INF;
dp[0][x] = 0;
for (int i = 1; i <= n; i++)
{
q.clear();
for (int j = 1; j <= m; j++)
{
if (q.size() && j - q.front().second > t) q.pop_front();
while (q.size() && q.back().first <= (dp[i - 1][j] - sum[i][j - 1]))
{
q.pop_back();
}
q.push_back({ dp[i - 1][j] - sum[i][j - 1] , j });
dp[i][j] = q.front().first + sum[i][j];
}
q.clear();
for (int j = m; j > 0; j--)
{
if (q.size() && q.front().second - j > t) q.pop_front();
while (q.size() && q.back().first <= (dp[i - 1][j] + sum[i][j]))
{
q.pop_back();
}
q.push_back({ dp[i - 1][j] + sum[i][j] , j });
dp[i][j] = max(dp[i][j], q.front().first - sum[i][j - 1]);
}
}
int maxx = -INF;
for (int i = 1; i <= m; i++) maxx = max(dp[n][i], maxx);
printf("%d
", maxx);
}
return 0;
}