| Time Limit: 2000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
uDebugDescription
Gappu has a very busy weekend ahead of him. Because, next weekend is Halloween, and he is planning to attend as many parties as he can. Since it's Halloween, these parties are all costume parties, Gappu always selects his costumes in such a way that it blends with his friends, that is, when he is attending the party, arranged by his comic-book-fan friends, he will go with the costume of Superman, but when the party is arranged contest-buddies, he would go with the costume of 'Chinese Postman'.
Since he is going to attend a number of parties on the Halloween night, and wear costumes accordingly, he will be changing his costumes a number of times. So, to make things a little easier, he may put on costumes one over another (that is he may wear the uniform for the postman, over the superman costume). Before each party he can take off some of the costumes, or wear a new one. That is, if he is wearing the Postman uniform over the Superman costume, and wants to go to a party in Superman costume, he can take off the Postman uniform, or he can wear a new Superman uniform. But, keep in mind that, Gappu doesn't like to wear dresses without cleaning them first, so, after taking off the Postman uniform, he cannot use that again in the Halloween night, if he needs the Postman costume again, he will have to use a new one. He can take off any number of costumes, and if he takes off k of the costumes, that will be the last k ones (e.g. if he wears costume A before costume B, to take off A, first he has to remove B).
Given the parties and the costumes, find the minimum number of costumes Gappu will need in the Halloween night.
Input
Input starts with an integer T (≤ 200), denoting the number of test cases.
Each case starts with a line containing an integer N (1 ≤ N ≤ 100) denoting the number of parties. Next line contains N integers, where the ith integer ci (1 ≤ ci ≤ 100) denotes the costume he will be wearing in party i. He will attend party 1 first, then party 2, and so on.
Output
For each case, print the case number and the minimum number of required costumes.
Sample Input
2
4
1 2 1 2
7
1 2 1 1 3 2 1
Sample Output
Case 1: 3
Case 2: 4
题意：给你n天需要穿的衣服的样式，每次可以套着穿衣服，脱掉的衣服就不能再用了（可以再穿），问至少要带多少条衣服才能参加所有宴会
http://lightoj.com/volume_showproblem.php?problem=1422
思路：首先dp[i][j]代表从区间i到区间j需要的最少穿衣服数量，我采取的是从下向上更新的。那么，面临第i件衣服，首先我们考虑穿上它，那么它所在的区间dp[i][j]=dp[i+1][j]+1；
接着考虑是否可以不用穿上它？在什么条件下，可以不用穿这件衣服呢？只有当区间i+1~~j里面已经穿过这件衣服的时候，就可以考虑不用再穿这件衣服。那么假设i+1<=k<=j
其中第k件衣服与第i件一样，那么第k件衣服穿上，第i件衣服不穿的情况的最少穿衣数==dp[i+1][k-1]+dp[k][j]；第i件衣服不穿，那么就从第i+1件衣服开始计算，第k件衣服存在，那么在这个断点，我们该怎么判断是dp[i+1][k-1]+dp[k][j]呢？可以直接从数据推导出这个关系，也可以这样，在第k区间的时候，a[k]可能不是它自己本身穿的，而是在第k区间到第j区间，存在一个k<tmp<=j，有a[tmp]==a[k]......所以会是dp[i+1][k-1]+dp[k][j]........
代码：#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#define INF 0x3fffffff
using namespace std;
int dp[105][105];
int a[105];
int main(){
int T;
scanf("%d",&T);
int t = 1;
while(T--){
int n;
scanf("%d",&n);
memset(dp,0,sizeof(dp));
for(int i = 1; i <= n; i++)
scanf("%d",&a[i]);
for(int i = 1; i <= n; i++)
dp[i][i] = 1;
for(int i = 2; i <= n; i++){
for(int j = 1; j <= n; j++){
dp[j][j+i-1] = 1 + dp[j][j+i-2];
int temp = INF;
int theend = j+i-1;
for(int k = theend -1; k >= j; k--){
if(a[k] == a[theend])temp = min(temp,dp[j][k-1]+dp[k][theend-1]);
}
dp[j][j+i-1] = min(temp,dp[j][j+i-1]);
}
}
printf("Case %d: %d\n",t++,dp[1][n]);
}
}