CodeForces 61E Enemy is weak 求i<j<k && a[i]>a[j]>a[k] 的對數 樹狀數組

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題目連結：點擊開啟連結

題意是求 i<j<k && a[i]>a[j]>a[k] 的對數

如果只有2元組那就是求逆序數的做法

三元組的話就用一個樹狀數組x表示 數字i前面有多少個比自己大的個數

然後每次給這個y數組求和，再把x中>a[i]的個數存入y中即可

#include <algorithm>#include <cctype>#include <cassert>#include <cstdio>#include <cstring>#include <climits>#include <vector>#include<iostream>using namespace std;#define ll long longinline void rd(int &ret){char c;do { c = getchar();} while(c < '0' || c > '9');ret = c - '0';while((c=getchar()) >= '0' && c <= '9')ret = ret * 10 + ( c - '0' );}#define N 1000005#define eps 1e-8#define inf 1000000ll n;struct node{ll c[N];inline ll lowbit(ll x){return x&-x;}void init(){memset(c, 0, sizeof c);}ll sum(ll x){ll ans = 0;while(x<=n+10)ans += c[x], x+=lowbit(x);return ans;}void change(ll x, ll y){while(x)c[x] +=y, x-=lowbit(x);}}x, y;int haifei[1000000], panting[1000000];int main(){ll i, j;while(cin>>n){ll ans = 0;for(i = 0; i < n; i++)rd(haifei[i]), panting[i] = haifei[i];x.init(); y.init();sort(haifei, haifei+n);for(i = 0; i < n; i++){ll b = (lower_bound(haifei, haifei+n, panting[i]) - haifei) +1;ll siz = y.sum(b);ans += siz;y.change(b, x.sum(b));x.change(b, 1);}cout<<ans<<endl;}return 0;}