LightOJ 1422 Halloween Costumes (区间DP)

题意：给你n天要穿的衣服，可以套着穿，但是一旦脱下来就不能再穿，问这n天最少需要准备多少件衣服。

析：dp[i][j] 表示第 i 天到 第 j 天最少要穿多少这衣服，对于第 i 天的衣服，可以是以后不用的那就是dp[i][j] = dp[i+1][j] + 1。

或者是第 k 又用,dp[i][j] = min(dp[i][j], dp[i+1][k-1]+dp[k][j])。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 100 + 5;
const LL mod = 1LL<<32;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int a[maxn];
int dp[maxn][maxn];

int main(){
  int T;  cin >> T;
  for(int kase = 1; kase <= T; ++kase){
    scanf("%d", &n);
    for(int i = 1; i <= n; ++i){
      scanf("%d", a+i);
      for(int j = i; j <= n; ++j)
        dp[i][j] = j - i + 1;
    }

    for(int i = n-1; i > 0; --i){
      for(int j = i+1; j <= n; ++j){
        dp[i][j] = dp[i+1][j] + 1;
        for(int k = i+1; k <= j; ++k)
          if(a[k] == a[i])  dp[i][j] = min(dp[i][j], dp[i+1][k-1]+dp[k][j]);
      }
    }
    printf("Case %d: %d\n", kase, dp[1][n]);
  }
  return 0;
}