2016 Multi-University Training Contest 5 1012 World is Exploding 树状数组+离线化

http://acm.hdu.edu.cn/showproblem.php?pid=5792

1012 World is Exploding

题意：选四个数，满足a<b and A[a]<A[b]   c<d and A[c]>A[d] 问有几个这样的集合

思路：

树状数组+离线化

先处理出每个数左边比它小 大，右边比它大 小的数目，用cnt[][i]表示。最后统计一下减去重复的就可以

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <sstream>
 #include <string>
 #include <algorithm>
 #include <list>
 #include <map>
 #include <vector>
 #include <queue>
 #include <stack>
 #include <cmath>
 #include <cstdlib>
 using namespace std;
 typedef long long ll;
 ll cnt[][];
 ll cnt1[],cnt2[];
 ll a[];

 struct Node{
     ll num,id;
 }sub_a[];

 ll maxn = ;
 ll n;
 ll tree[];
 bool cmp(Node a, Node b) { //按照数字排序
     return a.num < b.num;
 }
 void discrete() { //离散化
     sort(sub_a+, sub_a++n, cmp);
     a[sub_a[].id] = ;
     maxn = ;
     for(ll i = ; i <= n; i++) {
         if(sub_a[i].num != sub_a[i-].num)
             a[sub_a[i].id] = i;
         else
             a[sub_a[i].id] = a[sub_a[i-].id];
         maxn = max(maxn,a[sub_a[i].id]);
     }
 }
 void add(ll k){
     while(k <= n){
         tree[k] ++;
         k += k & (- k);
     }
 }
 ll read(ll k){
     ll ans = ;
     while(k){
         ans += tree[k];
         k -= k &(-k);
     }
     return ans;
 }

 int main(){
     while(scanf("%I64d",&n) != EOF)
     {
         memset(cnt1,,sizeof(cnt1));
         memset(cnt2,,sizeof(cnt2));
         for(ll i = ; i <= n; i ++){
             scanf("%I64d",&sub_a[i].num);
             sub_a[i].id = i;
         }
         discrete();
         memset(tree,,sizeof(tree));
         //在它右侧比它小的
         for(ll i = n; i >= ; i--){

             add(a[i]);
            cnt[][i] = read(a[i] - );
         }
         //在它左侧比它小的
          memset(tree,,sizeof(tree));
         for(ll i = ; i <= n; i ++){
             add(a[i]);
             cnt[][i] = read(a[i] - );
             a[i] = maxn +  - a[i];
         }
          //在它右侧比它大的
           memset(tree,,sizeof(tree));
         for(ll i = n; i >= ; i --){

             add(a[i]);
            cnt[][i] = read(a[i] - );
         }
         //在它左侧比它大的
          memset(tree,,sizeof(tree));
         for(ll i = ; i <= n; i ++){
             add(a[i]);
             cnt[][i] = read(a[i] - );
             a[i] = maxn +  - a[i];
         }
         ll cnta = ,cntb = ;
         for(ll i = ; i <= n; i ++){
             cnt1[i] = cnt[][i] + cnt[][i];
             cnta += cnt1[i];
             cnt2[i] = cnt[][i] + cnt[][i];
              cntb += cnt2[i];
         }
         cnta = cnta / ;
         cntb = cntb / ;
         long long  ans = cnta * cntb;
         for(ll i = ; i <= n; i ++){
             ans -= cnt1[i] * cnt2[i];
         }
         printf("%I64d\n",ans);
     }
     return ;
 }