4580: [Usaco2016 Open]248
Time Limit: 10 Sec
Memory Limit: 128 MB
Submit: 131
Solved: 105
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Description
Bessie likes downloading games to play on her cell phone, even though she does find the small touch
screen rather cumbersome to use with her large hooves.She is particularly intrigued by the current g
ame she is playing. The game starts with a sequence of NN positive integers (2≤N≤248), each in the
range 1…40. In one move, Bessie can take two adjacent numbers with equal values and replace them a
single number of value one greater (e.g., she might replace two adjacent 7s with an 8). The goal is
to maximize the value of the largest number present in the sequence at the end of the game. Please
help Bessie score as highly as possible!
Input
The first line of input contains N, and the next N lines give the sequence of N numbers at the start
of the game.
Output
Please output the largest integer Bessie can generate.
Sample Input
4
1
1
1
2
Sample Output
3
//In this example shown here, Bessie first merges the second and third 1s to obtain the sequence 1 2 2
, and then she merges the 2s into a 3. Note that it is not optimal to join the first two 1s.
HINT
Source
Gold
【分析】
签到题第三弹...
真的。没有1A的题，全都是2A，我今天脑子一定瓦特了。
用的是区间DP，不过我赌五毛贪心可以O(n)
【代码】
#include<iostream>
#include<cstring>
#include<cstdio>
#define ll long long
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=250;
int n,m,ans;
int dp[mxn][mxn];
inline int read()
{
int x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9') {if(ch=='-') f=-1;ch=getchar();}
while(ch>='0'&&ch<='9') x=(x<<1)+(x<<3)+ch-'0',ch=getchar();
return x*f;
}
int main()
{
int i,j,k;
n=read();
fo(i,1,n) dp[i][i]=read();
fo(k,1,n)
fo(i,1,n) if(i+k<=n)
fo(j,i,i+k-1)
{
if(dp[i][j]==dp[j+1][i+k])
dp[i][i+k]=max(dp[i][i+k],dp[i][j]+1);
ans=max(ans,dp[i][i+k]);
}
printf("%d\n",ans);
return 0;
}
/*
8
1 1 1 1 1 1 1 1
*/
