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 + ;

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

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

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

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