HDU-Yuna's confusion 树状数组 Or Multiset

比赛的时候竟然没有想到这题可以用树状数组过，由于数字的区间比较小，所以直接开设一个树状数组，定义sum(i) 表示小于i的数的总数。那么判定一个数是否有的条件就是计算sum(i+1) == sum(i) 便可以了，查找第K大的数，也就可以直接二分了。

代码如下：

#include <cstdlib>
#include <cstring>
#include <cstdio>
#include <algorithm>
#define MAXN 100000
using namespace std;int c[MAXN+5];int lowbit(int x)
{return x & -x;
}void modify(int x, int val)
{for (int i = x; i <= MAXN; i += lowbit(i)) {c[i] += val;}
}int sum(int x)
{int ret = 0;for (int i = x; i > 0; i -= lowbit(i)) {ret += c[i];}return ret;
}int bsearch(int l, int r, int k)
{int mid, ret = -1;while (l <= r) {mid = (l + r) >> 1;//    printf("mid = %d, sum() = %d\n", mid, sum(mid));//    getchar();if (sum(mid) >= k) {ret = mid;r = mid - 1;    }else {l = mid + 1;    }}return ret;
}int main()
{int M, x, y, ret;char op[5];while (scanf("%d", &M) == 1) {memset(c, 0, sizeof (c));for (int i = 1; i <= M; ++i) {scanf("%s", op);if (op[0] == 'A') {scanf("%d", &x);modify(x, 1);    }else if (op[0] == 'D'){scanf("%d", &x);//    printf("%d %d\n", sum(x), sum(x-1));if (sum(x) == sum(x-1)) {puts("No Elment!");}else {modify(x, -1);}}else {  // query通过二分查找来对付 scanf("%d %d", &x, &y);y += sum(x);ret = bsearch(x+1, MAXN, y);if (ret == -1) {puts("Not Find!");}else {printf("%d\n", ret);}}}    }return 0;    
}

还有一种做法，就是直接用STL中的multiset.

代码如下：

#include<iostream>
#include<algorithm>
#include<set>
using namespace std;multiset<int> s;
multiset<int>::iterator it;int main(){int Q;while( ~scanf("%d",&Q)){s.clear();char ope[10];int x,k;for(int i = 1; i <= Q; i++){scanf("%s", ope);if( ope[0] == 'A' ){scanf("%d",&x);s.insert(x);    }else if( ope[0] == 'D' ){scanf("%d",&x);int cnt = s.count(x);if( cnt == 0 ){puts("No Elment!");continue;    }s.erase(x);for(int i = 1; i <= cnt-1; i++)s.insert(x);    }else {scanf("%d%d",&x,&k);it = s.upper_bound(x);if( it == s.end() ){//|| (it+(k-1) >= s.end() ) ){puts("Not Find!");continue;}    int i=1;for(; it != s.end() && i < k; i++, it++); if( it == s.end() ){puts("Not Find!");continue;}printf("%d\n", *(it) );}}    }    
}