单调队列--P3800 Power收集

*传送

对于一个$n \times m$的矩形，我们只要对每一行求滑动窗口，当前位置的最大价值就是他上一行的滑动窗口（点击查看解释）+当前位置的价值。因为当前位置左上T个位置，会被他左边T个位置的点更新到，所以我们只需要算一边就好。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
const int maxn= 4005;
int N, M, K, T, val[maxn][maxn], f[maxn][maxn], g[maxn];
int head, tail, q[maxn];
int main() {
    cin >> N >> M >> K >> T;
    for (register int i = 1; i <= K; ++i) {
        int x, y, z;
        scanf("%d%d%d", &x, &y, &z);
        val[x][y] = z;
    }
    memset(f, -0x3f, sizeof(f));
    for (int i = 1; i <= M; ++i) 
        f[1][i] = val[1][i];
    tail = 0, head = 1;
    for (register int i = 2; i <= N; ++i) {
        q[head = 0] = q[tail = 1] = 0;
        for (register int j = 1; j <= M; ++j) {            
            while (head <= tail && j - q[head]>T) head++;
            while (head <= tail && f[i - 1][q[tail]] <= f[i - 1][j]) tail--;
            q[++tail] = j, g[j] = f[i - 1][q[head]];
        }
          for (register int j = 1; j <= M; ++j) {
              int l = j, r = min(j + T, M);
              f[i][j] = max(g[l], g[r]) + val[i][j];
        }
    }
    int Ans = -2e9;
    for (int i = 1; i <= M; ++i) 
        Ans = max(Ans, f[N][i]);
    printf("%d\n", Ans);
    return 0;
}