[luoguP2736] “破锣摇滚”乐队 Raucous Rockers（DP）

传送门

f[i][j]表示前i首歌放到前j个盘里最多能放多首

ntr[i][j]表示i~j中最多能放进一张盘中多少首歌 

ntr数组可以贪心预处理出来。

#include <cstdio>
#include <iostream>
#include <algorithm>
#define N 21 
#define max(x, y) ((x) > (y) ? (x) : (y))

int n, t, m;
int a[N], b[N], ntr[N][N], f[N][N];
//f[i][j]表示前i首歌放到前j个盘里最多能放多首
//ntr[i][j]表示i~j中最多能放进一张盘中多少首歌 

inline int read()
{
	int x = 0, f = 1;
	char ch = getchar();
	for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = -1;
	for(; isdigit(ch); ch = getchar()) x = (x << 1) + (x << 3) + ch - '0';
	return x * f;
}

inline void init()
{
	int i, j, k, sum, cnt;
	for(i = 1; i <= n; i++)
		for(j = i; j <= n; j++)
		{
			sum = cnt = 0;
			for(k = i; k <= j; k++) b[k] = a[k];
			std::sort(b + i, b + j + 1);
			for(k = i; k <= j; k++)
				if(sum + b[k] <= t)
				{
					cnt++;
					sum += b[k];
				}
			ntr[i][j] = cnt;
		}
}

int main()
{
	int i, j, k;
	n = read();
	t = read();
	m = read();
	for(i = 1; i <= n; i++) a[i] = read();
	init();
	for(i = 1; i <= n; i++)
		for(j = 1; j <= m; j++)
			for(k = 0; k < i; k++)
				f[i][j] = max(f[i][j], f[k][j - 1] + ntr[k + 1][i]);
	printf("%d
", f[n][m]);
	return 0;
}