标签:tar std www pid auc get ++ print ini
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\n", f[n][m]);
return 0;
}
[luoguP2736] “破锣摇滚”乐队 Raucous Rockers(DP)
标签:tar std www pid auc get ++ print ini
原文地址:http://www.cnblogs.com/zhenghaotian/p/7337253.html