nyoj117求逆序数(离散化+树状数组/归并排序)

题目链接：nyoj117

树状数组思路：

首先将输入的数组离散化，使各个元素比较接近，而不是离散的。

离散时用一个结构体，val表示原来输入的数，pos表示下标；接着对结构体按val的大小排序，此时，val和结构体的下标就是一个一一对应的关系，而且满足原来的大小关系。然后用数组reflect存储原来所有的大小信息。

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int N = 1000005;int reflect[N],c[N],n;struct node{    int val;    int pos;}s[N];bool cmp(node x,node y){    if(x.val == y.val) return x.pos < y.pos;    return x.val < y.val;}int lowbit(int x){    return x&(-x);}void update(int x){    while(x <= n)    {        c[x] += 1;        x += lowbit(x);    }}long long getsum(int x)//统计比x小的数{    long long sum = 0;    while(x > 0)    {        sum += c[x];        x -= lowbit(x);    }    return sum;}int main(){int T,i;scanf("%d",&T);while(T--){    scanf("%d",&n);    for(i = 1; i <= n; i ++)            scanf("%d",&s[i].val), s[i].pos = i;        sort(s+1,s+n+1,cmp);        for(i = 1; i <= n; i ++)            reflect[s[i].pos] = i;        memset(c,0,sizeof(c));        long long ans = 0;        for(i = 1; i <= n; i ++)        {            update(reflect[i]);            ans += i - getsum(reflect[i]);//i表示已经插入的数的个数        }        printf("%lld\n",ans);}return 0;}

归并排序：

#include<stdio.h>int a[1000005],temp[1000005];long long sum;void merge(int left,int mid,int right){int i = left,j = mid + 1;int k = left;while(i <=mid && j <= right){if(a[i] <= a[j])temp[k ++] = a[i ++];else{temp[k ++] = a[j ++];sum += mid - i + 1;//关键}}while(i <= mid)        temp[k ++] = a[i ++];while(j <= right)        temp[k ++] = a[j ++];for(i = left ; i <= right; i ++)a[i] = temp[i];}void msort(int left,int right){if(left < right){int mid = (left + right) >> 1;msort(left,mid);msort(mid + 1,right);merge(left,mid,right);}}int main(){int i,n,T;scanf("%d",&T);while(T--){    scanf("%d",&n);for(i = 0;i < n; i ++)scanf("%d",&a[i]);        sum = 0;msort(0,n - 1);printf("%lld\n",sum);}return 0;}