sdut 1309 —— 不老的传说问题

题目： http://acm.sdut.edu.cn/sdutoj/problem.php?action=showproblem&problemid=1309

状态：dp[i][j] := 区间i~j都是无色时的最小涂色次数

转移：dp[i][j] = min(dp[i+1][j]+1 (单独涂色), dp[i+1][k]+dp[k+1][j] (借助它人涂色))  (i<k<=min(j,i+K-1) && a[i]==a[k])

#include <cstdio>
#include <iostream>

using namespace std;

const int MAXN = 200*2+5;
int a[MAXN];
int dp[MAXN][MAXN];

int main ()
{
    int N, C, K;
    while(scanf("%d%d%d", &N, &C, &K) != EOF) {
        for(int i=0; i<N; i++) {
            scanf("%d", &a[i]);
            a[N+i] = a[i];
        }
        for(int step=0; step<N; step++) {
            for(int i=0; i+step<2*N; i++) {
                int j = i + step;
                dp[i][j] = dp[i+1][j] + 1;
                int t = min(j, i+K-1);
                for(int k=i+1; k<=t; k++) {
                    if(a[k]==a[i]) {
                        dp[i][j] = min(dp[i][j], dp[i+1][k] + dp[k+1][j]);
                    }
                }
            }
        }
        int ans = N;
        for(int i=0; i<N; i++) {
            ans = min(ans, dp[i][i+N-1]);
        }
        printf("%d\n", ans);
    }
    return 0;
}