hdu6212[区间dp] 2017青岛ACM-ICPC网络赛

原题: BZOJ1032 （原题数据有问题）

/*hdu6212[区间dp] 2017青岛ACM-ICPC网络赛*/
#include <bits/stdc++.h>
using namespace std;
int T, n, tot = , kase = ;
char s[];
int dp[][], cnt[], col[];
void solve() {
    memset(dp, 0x3f, sizeof(dp));
    tot = ;
    n = strlen(s);
    cnt[tot] = , col[tot] = s[] - '';
    for (int i = ; i < n; i++) {
        if (s[i] == s[i - ]) cnt[tot]++;
        else {
            cnt[++tot] = ;
            col[tot] = s[i] - '';
        }
    }
    n = tot;
    for (int i = ; i <= n; i++) dp[i][] = cnt[i] >=  ?  : ;
    for (int j = ; j <= n; j++) {
        for (int i = ; i + j -  <= n; i++) {
            if (col[i] == col[i + j - ]) {
                if (cnt[i] + cnt[i + j - ] < ) {
                    dp[i][j] = dp[i + ][j - ] + (cnt[i] + cnt[i + j - ] == );
                    for (int k = ; k < j; k++) {
                        if (col[i + k - ] == col[i] && cnt[i + k - ] == ) {
                            dp[i][j] = min(dp[i][j], dp[i + ][k - ] + dp[i + k][j - k - ]);
                        }
                    }
                }
                else dp[i][j] = dp[i + ][j - ];
            }
            for (int k = ; k < j; k++) {
                dp[i][j] = min(dp[i][j], dp[i][k] + dp[i + k][j - k]);
            }
        }
    }
    printf("Case #%d: %d\n", kase++, dp[][n]);
}
int main() {
    scanf("%d", &T);
    while (T--) {
        scanf("%s", s);
        solve();
    }
    return ;
}
// 011011100000001010101110011000 ans=4
// 10110011001101  ans=3