【BZOJ3211】【并查集+树状数组】花神游历各国

Description

 

Input

 

Output

每次x=1时，每行一个整数，表示这次旅行的开心度

Sample Input

4

1 100 5 5

5

1 1 2

2 1 2

1 1 2

2 2 3

1 1 4

Sample Output

101

11

11

【分析】

开始看一眼觉得线段树可做。

后来看题解用树状数组瞬秒......orzzz，注意到任何一个int都可以在很小的次数下变为1，所以直接暴力单点修改，将变成1的数parent设为它右边的数。

注意，输入中可能会有很多的0....直接写优化不好会T。

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <cstring>
 #include <vector>
 #include <utility>
 #include <iomanip>
 #include <string>
 #include <cmath>
 #include <queue>
 #include <map>

 const int MAXN =  + ;
 const int MAX =  + ;
 using namespace std;
 typedef long long ll;
 int n;
 ll C[MAXN];
 int data[MAXN], parent[MAXN];

 int SQRT(int x){return (int)floor(sqrt((double)x * 1.0));}
 int lowbit(int x){return x & -x;}
 void add(int x, int val){
      while (x <= n){
            C[x] += val;
            x += lowbit(x);
      }
      return;
 }
 ll sum(int x){
     ll cnt = ;
     while (x > ){
           cnt += C[x];
           x -= lowbit(x);
     }
     return cnt;
 }
 int find(int x) {return parent[x] == x? x : parent[x] = find(parent[x]);}
 void init(){
      memset(C, , sizeof(C));
      scanf("%d", &n);
      for (int i = ; i <= n; i++){
          scanf("%d", &data[i]);
          add(i, data[i]);
          parent[i] = i;
          if (data[i] ==  || data[i] == ) parent[i] = i + ;
      }
      parent[n + ] = n + ;
 }
 void work(){
      int q;
      scanf("%d", &q);
      for (int i = ; i <= q; i++){
          int t, x, y;
          scanf("%d%d%d", &t, &x, &y);
          if (t == ) printf("%lld\n", sum(y) - sum(x - ));
          else{
               for (int i = x; i <= y; i = find(i + )){
                   int tmp = data[i];
                   data[i] = SQRT(data[i]);
                   add(i, data[i] - tmp);
                   if (data[i] <= ) parent[i] = find(i + );
               }
          }
      }
 }

 int main(){
     int T;
     #ifdef LOCAL
     freopen("data.txt",  "r",  stdin);
     freopen("out.txt",  "w",  stdout);
     #endif
     init();
     work();
     return ;
 }