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.
Examples 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.
題目鏈接
題意:有一段數字串,每次可以消除一段回文串,問最后將這段數字串完全消除至少要幾次。
解題思路:區間dp,如果st[i]==st[j],那么分兩種情況,len=2的時候,dp[i][j]=1,len>2的時候,dp[i][j]=dp[i+1][j-1]. 然后我們得到轉移方程,dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]). 最后dp[0][n-1]就為這個數字串需要消除的次數。
#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int dp[505][505],st[505];int n;int main(){ scanf("%d",&n); for(int i=0;i<n;i++) scanf("%d",&st[i]); for(int i=0;i<n;i++){ for(int j=i;j<n;j++) dp[i][j]=9999; } for(int i=0;i<n;i++){ dp[i][i]=1; } for(int l=2;l<=n;l++){ for(int i=0;i+l-1<n;i++){ int j=i+l-1; if(st[i]==st[j]&&l>2) dp[i][j]=dp[i+1][j-1]; else if(st[i]==st[j]) dp[i][j]=1; for(int k=i;k<j;k++){ dp[i][j]=min(dp[i][j],dp[i][k]+dp[k+1][j]); } } } printf("%d/n",dp[0][n-1]); return 0;}新聞熱點
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