hdu 3237

dp 状态压缩

#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <map>
#include <set>
#include <queue>
#include <stack>
#include <vector>
#include <sstream>
#include <string>
#include <cstring>
#include <algorithm>
#include <iostream>
#define maxn 105
#define INF 0x3f3f3f3f
#define inf 10000000
#define MOD 100000000
#define ULL unsigned long long
#define LL long long

using namespace std;

int hi[maxn], dp[2][maxn][1<<9][10], n, m, one[1<<9], mh, begin;

int countone(int x) {
    int ans = 0;
    for(int i = 0; i < 8; ++ i) 
        if(x&(1<<i)) ans ++;
    return ans;
}

void init() {
    begin = mh = 0;
    for(int i = 0; i < (1 << 8); ++ i) {
        one[i] = countone(i);
    }
}

int main()
{
    int ca = 0;
    init();
    while(scanf("%d%d", &n, &m) == 2 && n+m) {
        // printf("ff: %d\n", num);
        begin = mh = 0;
        for(int i = 0; i < n; ++ i) {
            scanf("%d", &hi[i]);
            hi[i] -= 25;
            mh = max(hi[i], mh);
            begin |= (1 << hi[i]);
        }
        mh ++;
        int tot = 1<<mh;
        for(int i = 0; i <= m; ++ i) {
            for(int j = 0; j < tot; ++ j) {
                for(int k = 0; k <= mh; ++ k) {
                    dp[0][i][j][k] = INF;
                }
            }
        }

        dp[0][0][1<<hi[0]][hi[0]] = 1;
        dp[0][1][0][mh] = 0;
        int now, pre;
        for(int i = 1; i < n; ++ i) {
            now = i%2;
            pre = 1-now;
            
            for(int j = 0; j <= m && j <= i+1; ++ j) {
                for(int k = 0; k < tot; ++ k) {
                    for(int q = 0; q <= mh; ++ q) {
                        dp[now][j][k][q] = INF;
                    }
                }
            }

            for(int j = 0; j <=  m && j <= i; ++ j) {
                for(int k = 0; k < tot; ++ k) {
                    for(int q = 0; q <= mh; ++ q) {
                        if(dp[pre][j][k][q] == INF) continue;
                        int nowk = k|(1<<hi[i]);
                        if(j < m) dp[now][j+1][k][q] = min(dp[now][j+1][k][q], dp[pre][j][k][q]);
                        if(hi[i] == q) {
                            dp[now][j][k][q] = min(dp[now][j][k][q], dp[pre][j][k][q]);
                        }
                        else {
                            dp[now][j][nowk][hi[i]] = min(dp[now][j][nowk][hi[i]], dp[pre][j][k][q]+1);
                        }
                    }
                }
            }
        }

        int ans = n;
        for(int i = 0; i <= m; ++ i) {
            for(int j = 0; j < tot; ++ j) {
                for(int k = 0; k < mh; ++ k) {
                    int st = begin^j;
                    ans = min(ans, one[st]+dp[now][i][j][k]);
                }
            }
        }

        printf("Case %d: %d\n\n", ++ca, ans);
    }
    return 0;
}