P3146 [USACO16OPEN]248 G(区间DP)
作者:互联网
题目描述
Bessie likes downloading games to play on her cell phone, even though she doesfind the small touch screen rather cumbersome to use with her large hooves.
She is particularly intrigued by the current game she is playing.The game starts with a sequence of NN positive integers (2≤N≤2482≤N≤248), each in the range 1…401…40. In one move, Bessie cantake two adjacent numbers with equal values and replace them a singlenumber of value one greater (e.g., she might replace two adjacent 7swith an 8). The goal is to maximize the value of the largest numberpresent in the sequence at the end of the game. Please help Bessiescore as highly as possible!
给定一个1*n的地图,在里面玩2048,每次可以合并相邻两个(数值范围1-40),问序列中出现的最大数字的值最大是多少。注意合并后的数值并非加倍而是+1,例如2与2合并后的数值为3。
输入格式
The first line of input contains NN, and the next NN lines give the sequence
of NN numbers at the start of the game.
输出格式
Please output the largest integer Bessie can generate.
输入输出样例
输入 #1复制
4
1
1
1
2
输出 #1复制
3
区间dp典型例题。注意如果区间不能完全合并也要标记,否则会T飞。
#include <bits/stdc++.h>
using namespace std;
int n, a[105], dp[405][405];//dp[i][j]表示第i个到第j个这一段全部合并能够得到的最大值
int f(int l, int r) {
if(dp[l][r] != 0) return dp[l][r];
for(int i = l; i < r; i++) {
if(f(l, i) == -1 || f(i + 1, r) == -1) continue;//如果当前段不能全部合并 也要标记
dp[l][r] = max(dp[l][r], (f(l, i) == f(i + 1, r) ? f(l, i) + 1 : 0));
}
if(dp[l][r] == 0) dp[l][r] = -1;
return dp[l][r];
}
int main() {
cin >> n;
memset(dp, 0, sizeof(dp));
for(int i = 1; i <= n; i++) {
cin >> a[i];
dp[i][i] = a[i];
}
int ans = 0;
for(int i = 1; i <= n; i++) {
for(int j = 1; j <= n; j++) {
ans = max(ans, f(i, j));
}
}
cout << ans << endl;
return 0;
}
标签:NN,int,she,合并,game,DP,248,dp,USACO16OPEN 来源: https://www.cnblogs.com/lipoicyclic/p/15428164.html