SGU 180 Inversions

点击打开sgu 180

思路: 树状数组+离散化

分析:

1 题目给定n个数，每个数最大为10^9 , 最小为0，求多少个匹配使得1<=i<j<=n并且A[i] > A[j]

2 因为数的最大值为10^9.所以我们应该利用离散化的思想。然后在利用树状数组求个数即可

3 注意由于输入的值由可能为0，我们应该在输入的时候就把所有的值加上1

代码:

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;

const int MAXN = 66000;

int n;
int num[MAXN];
int treeNum[MAXN];

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

int getSum(int x){
int sum = 0;
while(x){
sum += treeNum[x];
x -= lowbit(x);
}
return sum;
}

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

int search(int x){
int left = 0;
int right = n-1;
while(left <= right){
int mid = (left+right)>>1;
if(num[mid] == x)
return mid+1;
else if(num[mid] < x)
left = mid+1;
else
right = mid-1;
}
}

long long solve(){
int tmp[MAXN];
memcpy(tmp , num , sizeof(num));
sort(num , num+n);
long long ans = 0;
for(int i = 0 ; i < n ; i++){
int x = search(tmp[i]);
ans += i-getSum(x);
add(x , 1);
}
return ans;
}

int main(){
int x;
while(scanf("%d" , &n) != EOF){
memset(treeNum , 0 , sizeof(treeNum));
for(int i = 0 ; i < n ; i++){
scanf("%d" , &num[i]);
num[i]++;
}
printf("%lld\n" , solve());
}
return 0;
}