Codeforces Round #336 (Div. 2) D. Zuma 区间dp
2015-12-24 16:43
253 查看
[b]D. Zuma[/b]
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 ≤ n ≤ 500) — the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 ≤ ci ≤ n) — the color of the i-th gemstone in a line.
Output
Print a single integer — the minimum number of seconds needed to destroy the entire line.
Sample test(s)
input
output
input
output
input
output
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
题意:给你一个串,每次可以消除 其一个回文子串,问你最少消除几次
题解:设dp[l][r] 为 l到r 需要消除几次 ,
搜索就好
代码
Genos recently installed the game Zuma on his phone. In Zuma there exists a line of n gemstones, the i-th of which has color ci. The goal of the game is to destroy all the gemstones in the line as quickly as possible.
In one second, Genos is able to choose exactly one continuous substring of colored gemstones that is a palindrome and remove it from the line. After the substring is removed, the remaining gemstones shift to form a solid line again. What is the minimum number of seconds needed to destroy the entire line?
Let us remind, that the string (or substring) is called palindrome, if it reads same backwards or forward. In our case this means the color of the first gemstone is equal to the color of the last one, the color of the second gemstone is equal to the color of the next to last and so on.
Input
The first line of input contains a single integer n (1 ≤ n ≤ 500) — the number of gemstones.
The second line contains n space-separated integers, the i-th of which is ci (1 ≤ ci ≤ n) — the color of the i-th gemstone in a line.
Output
Print a single integer — the minimum number of seconds needed to destroy the entire line.
Sample test(s)
input
3 1 2 1
output
1
input
3 1 2 3
output
3
input
7 1 4 4 2 3 2 1
output
2
Note
In the first sample, Genos can destroy the entire line in one second.
In the second sample, Genos can only destroy one gemstone at a time, so destroying three gemstones takes three seconds.
In the third sample, to achieve the optimal time of two seconds, destroy palindrome 4 4 first and then destroy palindrome 1 2 3 2 1.
题意:给你一个串,每次可以消除 其一个回文子串,问你最少消除几次
题解:设dp[l][r] 为 l到r 需要消除几次 ,
搜索就好
//meek///#include<bits/stdc++.h> #include <iostream> #include <cstdio> #include <cmath> #include <string> #include <cstring> #include <algorithm> #include <queue> #include <map> #include <set> #include <stack> #include <sstream> #include <vector> using namespace std ; #define mem(a) memset(a,0,sizeof(a)) #define pb push_back #define fi first #define se second #define MP make_pair typedef long long ll; const int N = 510; const int inf = 99999999; const int mod= 1000000007; int dp ,a ,n; int dfs(int l,int r) { if(dp[l][r] != -1) return dp[l][r]; int& ret=dp[l][r]=inf; if(l==r) ret=1; else if(r-l == 1) { if(a[l]==a[r]) ret = 1; else ret = 2; } else { if(a[l] == a[r]) ret = dfs(l+1,r-1); for(int i=l;i<r;i++) ret = min(ret,dfs(l,i)+dfs(i+1,r)); } return ret; } int main() { scanf("%d",&n); memset(dp,-1,sizeof(dp)); for(int i=1;i<=n;i++) { scanf("%d",&a[i]); } printf("%d\n",dfs(1,n)); return 0; }
代码
相关文章推荐
- android中使用百度地图绘制弹出框的覆盖物
- note1
- php添加拓展
- Gradle 读书笔记(一)
- 浅谈千万级PV/IP规模高性能高并发网站架构
- IIC时序详解
- 《android开发必知的50个诀窍》笔记Hack-1
- 获取当前匹配元素 包括自身的html
- JaveScript RegExp的用法
- Delphi中调用API函数经验点滴(三)
- Android之判断某个服务是否正在运行的方法
- JavaScript求最大数最小数
- (ARM v7)自旋锁、读写锁、顺序锁代码追踪
- js获取当前url信息
- 微信 ua
- js常用正则表达式
- iOS 匆忙第二篇 — 页面传值
- iOS在ARC下保留dealloc的原因
- Android下的so库实例
- JAVA基础(14) JSTL的C标签使用详解