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| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 53777 | Accepted: 19766 |
Description
In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence Input
Output
Sample Input
5 9 1 0 5 4 3 1 2 3 0
Sample Output
6 0
Source
题意:求解一个串的逆序数的个数是多少？？
题解:离散化数组变成下标，然后每次将离散化的下标放进树状数组，放进去之后统计小于他的数的个数是多少。用 i - getsum(a[i])即为大于它的数的个数，其中 i 为当前已经插入的数的个数。
#include <iostream> #include <stdio.h> #include <string.h> #include <stack> #include <vector> #include <algorithm> using namespace std; const int N = 500005; struct Node{ int v,id; }node[N]; int a[N],c[N],n; int lowbit(int x){ return x&(-x); } void update(int idx,int v){ for(int i=idx;i<=N;i+=lowbit(i)){ c[i]+=v; } } int getsum(int idx){ int sum = 0; for(int i=idx;i>=1;i-=lowbit(i)){ sum+=c[i]; } return sum; } int cmp(Node a,Node b){ return a.v<b.v; } int main() { while(scanf("%d",&n)!=EOF,n){ memset(c,0,sizeof(c)); for(int i=1;i<=n;i++){ scanf("%d",&node[i].v); node[i].id = i; } sort(node+1,node+n+1,cmp); for(int i=1;i<=n;i++){ a[node[i].id] = i; } long long cnt = 0; for(int i=1;i<=n;i++){ update(a[i],1); cnt=cnt+ i - getsum(a[i]); } printf("%lld\n",cnt); } return 0; }
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原文地址：http://www.cnblogs.com/liyinggang/p/5655293.html
