Luogu1070-道路游戏-动态规划

Solution

用对角线的前缀和快速进行转移，复杂度$O(N^3)$， 洛谷神机太快了$N^3$都能过

然而正解是单调队列优化， 能优化到$O(N^2)$，然而我弱得什么都不会

 

Code

#include<cstdio>
#include<cstring>
#include<algorithm>
#define rd read()
#define rep(i,a,b) for(register int i = (a); i <= (b); ++i)
#define per(i,a,b) for(register int i = (a); i >= (b); --i)
using namespace std;

const int N = 2e3 + 5;

int n, m, p;
int a[N][N], sum[N][N], cost[N];
int f[N], ans;

inline int read() {
    int X = 0, p = 1; char c = getchar();
    for(; c > '9' || c < '0'; c = getchar()) if(c == '-') p = -1;
    for(; c >= '0' && c <= '9'; c = getchar()) X = X * 10 + c - '0';
    return X * p;
}

inline int ch(int x) {
    return (x - 1 + n) % n + 1;
}

int main()
{
    n = rd; m = rd; p = rd;
    rep(i, 1, n) rep(j, 1, m) a[i][j] = rd;
    rep(j, 1, m) rep(i, 1, n) sum[i][j] = sum[ch(i - 1)][j - 1] + a[i][j];
    rep(i, 1, n) cost[i] = rd;
    memset(f, 128, sizeof(f));
    f[1] = 0;
    rep(j, 1, m) rep(i, 1, n) rep(k, 1, p) {
        if(j + k > m + 1) break;
        f[j + k] = max(f[j + k], f[j] + sum[ch(i + k - 1)][j + k - 1] - sum[ch(i - 1)][j - 1] - cost[i]);
    }
    rep(i, 1, m + 1) ans = max(ans, f[i]);
    printf("%d\n", ans);
}