UVa 1513 Movie collection (树状数组)

题意：给定给你一叠DV，编号1到n，1在最上面，n在最下面。然后现在给你m个操作，每次都指定一张CD，问要拿走这个CD需要挪走上面多少张CD，并且这个要拿走的CD放在这个叠CD的顶端。

析：这个题要倒着来做，我就是正着做的，太复杂了，因为每次操作后，都要再重新处理后面的数，时间复杂度太高。

如果是倒着做，每次只要更新后面的就好，不用管前面的数，因为前面的是不会影响到的。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 300000 + 10;
const int mod = 100000 + 50;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r > 0 && r <= n && c > 0 && c <= m;
}
int sum[maxn];

int lowbit(int x){ return -x&x; }

void add(int x, int val){
  while(x < maxn){
    sum[x] += val;
    x += lowbit(x);
  }
}

int query(int x){
  int ans = 0;
  while(x){
    ans += sum[x];
    x -= lowbit(x);
  }
  return ans;
}
int pos[maxn];

int main(){
  int T;  cin >> T;
  while(T--){
    scanf("%d %d", &n, &m);
    memset(sum, 0, sizeof sum);
    for(int i = 1; i <= n; ++i){
      pos[i] = n - i + 1;
      add(pos[i], 1);
    }

    int cnt = n;
    for(int i = 0; i < m; ++i){
      if(i)  putchar(‘ ‘);
      int x;
      scanf("%d", &x);
      printf("%d", n - query(pos[x]));
      add(pos[x], -1);
      pos[x] = ++cnt;
      add(pos[x], 1);
    }
    printf("\n");
  }
  return 0;
}