CodeForcesGym 100524J Jingles of a String

Jingles of a String

Time Limit: 2000ms

Memory Limit: 524288KB

This problem will be judged on CodeForcesGym. Original ID: 100524J
64-bit integer IO format: %I64d      Java class name: (Any)

 

 

解题：题目很厉害啊！借鉴某大牛的思路。

 

由于都是小写字母，所以对任意一个区间，最多只有26个，用二进制表示该区间出现过的字母

 

然后就是固定一个左边界，可以算出左边界到字符串末这区间中1个个数，那么枚举这些1的个数，进行二分查找出现这么多个1的区间的最大右边界。

 

RMQ st可以这么用，也是涨姿势了

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 100005;
int st[maxn][20];
char str[maxn];
int query(int L,int R) {
    int lg = 31 - __builtin_clz(R - L + 1);
    return st[L][lg]|st[R - (1<<lg) + 1][lg];
}
unordered_map<int,int>ump;
int main() {
#define NAME "jingles"
    freopen(NAME".in","r",stdin);
    freopen(NAME".out","w",stdout);
    int kase;
    scanf("%d",&kase);
    while(kase--) {
        scanf("%s",str);
        int len = strlen(str);
        for(int i = 0; i < len; ++i)
            st[i][0] = 1<<(str[i] - 'a');
        for(int j = 1; (1<<j) <= len; ++j) {
            for(int i = 0; i + (1<<j) <= len; ++i)
                st[i][j] = st[i][j-1]|st[i+(1<<(j-1))][j-1];
        }
        ump.clear();
        for(int i = 0; i < len; ++i) {
            int x = query(i,len - 1);
            ump[x] = max(ump[x],len - i);
            int cnt = __builtin_popcount(x);
            for(int j = 1; j < cnt; ++j) {
                int low = i,high = len-1,ret;
                while(low <= high) {
                    int mid = (low + high)>>1;
                    int y = __builtin_popcount(query(i,mid));
                    if(y <= j) {
                        low = mid + 1;
                        ret = mid;
                    } else high = mid - 1;
                }
                ump[query(i,ret)] = max(ump[query(i,ret)],ret - i + 1);
            }
        }
        LL ans = 0;
        for(auto &it:ump)
            ans += __builtin_popcount(it.first)*(LL)it.second;
        printf("%d %I64d
",ump.size(),ans);
    }
    return 0;
}