ZOJ - 3635 Cinema in Akiba（树状数组+二分）

Posted 2020-09-07 SomnusMistletoe

题意：已知有n个人，从第一个人开始每个人被安排在第ai个空座上，有m组询问，问某人所坐的位置。

分析：

1、用树状数组维护空座的个数，方法：

将所有的空座初始化为1，sum（x）则表示从座位1到座位x空座的个数。

2、对于每个人，根据sum(mid)，二分找使sum(mid)大于等于a[i]的最小的mid，即第ai个空座的位置，并将该位置加上-1，则该位置的值变为0，从而不参与空座数的统计。

3、vis[q]即为标号为q的人所坐的位置。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define lowbit(x) (x & (-x))
const double eps = 1e-8;
inline int dcmp(double a, double b){
    if(fabs(a - b) < eps) return 0;
    return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 50000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
int vis[MAXN];
int a[MAXN];
int z[MAXN];
int n;
int sum(int x){
    int ans = 0;
    for(int i = x; i >= 1; i -= lowbit(i)){
        ans += z[i];
    }
    return ans;
}
void add(int x, int value){
    for(int i = x; i <= n; i += lowbit(i)){
        z[i] += value;
    }
}
int solve(int x){
    int l = 1, r = n;
    while(l < r){
        int mid = l + (r - l) / 2;
        if(sum(mid) >= x) r = mid;
        else l = mid + 1;
    }
    return r;
}
int main(){
    while(scanf("%d", &n) == 1){
        memset(vis, 0, sizeof vis);
        for(int i = 1; i <= n; ++i){
            scanf("%d", &a[i]);
            add(i, 1);
        }
        for(int i = 1; i <= n; ++i){
            vis[i] = solve(a[i]);
            add(vis[i], -1);
        }
        int m;
        scanf("%d", &m);
        bool flag = true;
        while(m--){
            int q;
            scanf("%d", &q);
            if(flag) flag = false;
            else printf(" ");
            printf("%d", vis[q]);
        }
        printf("\n");
    }
    return 0;
}