ZOJ 4117 BaoBao Loves Reading (樹狀數組 前綴和)

題目鏈接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=6019

用樹狀數組維護書桌整個時間段的書本數目，區間查詢得到出現相同書本時的時間差，即爲可以省去的去書櫃拿書次數(書桌容量爲即時間差)，最後利用前綴和，得到容量1~n的可省次數

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <vector>
#include <string>
#include <cstring>
#include <queue>
#define eps 1e-8
using namespace std;
typedef long long ll;
static const int MAX_N = 1e5 + 5;
static const ll Mod = 233;
static const int INF = 0x3f3f3f3f;
int C[MAX_N], pr[MAX_N], rev[MAX_N];
int lowbit(int x){
    return x & (-x);
}
void add(int x, int y, int n){
    for(int i = x; i <= n; i += lowbit(i))C[i] += y;
}
int get_sum(int x, int y){
    int ret1 = 0, ret2 = 0;
    for(int i = y; i; i -= lowbit(i)) ret2 += C[i];
    for(int i = x; i; i -= lowbit(i)) ret1 += C[i];
    return ret2 - ret1;
}
int main(){
//    freopen("input.txt", "r", stdin);
//    freopen("output.txt", "w", stdout);
    int T;
    scanf("%d", &T);
    while(T--){
        int n;
        scanf("%d", &n);
        for(int i = 0; i <= n; ++i){
            C[i] = 0;
            pr[i] = 0;
            rev[i] = 0;
        }
        for(int i = 1; i <= n; ++i){
            int v;
            scanf("%d", &v);
            if(pr[v]){
                add(pr[v], -1, n);
                add(i, 1, n);
                int p = get_sum(pr[v] - 1, i);    
                ++rev[p];    //容量爲p時可省次數+1
            }
            else add(i, 1, n);
            pr[v] = i;
        }
        for(int i = 1; i <= n; ++i) rev[i] += rev[i - 1];    //前綴和,前面能省的，後面一定能省
        for(int i = 1; i <= n; ++i) printf("%d%c", n - rev[i], i == n ? '\n' : ' ');
    }
}