Inversion
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 914 Accepted Submission(s): 380
Problem Description
bobo has a sequence a1,a2,…,an. He is allowed to swap two adjacent numbers for no more than k times.
Find the minimum number of inversions after his swaps.
Note: The number of inversions is the number of pair (i,j) where 1≤i<j≤n and ai>aj.
Input
The input consists of several tests. For each tests:
The first line contains 2 integers n,k (1≤n≤105,0≤k≤109). The second line contains n integers a1,a2,…,an (0≤ai≤109).
Output
For each tests:
A single integer denotes the minimum number of inversions.
Sample Input
3 1 2 2 1 3 0 2 2 1
Sample Output
1 2
题意：n个数，最多有k次相邻位置的数的交换，问最小的逆序数为多少
思路：保证每次交换逆序数都减小，可以参考冒泡排序的做法。最后只要计算原数列逆序数，然后减掉k即可。
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
#include <map>
using namespace std;
const int maxn = 100000+10;
int sum[maxn];
int n,k;
int num[maxn],tn[maxn];
map<int,int> mma;
bool cmp(int a,int b){
return a > b;
}
int lowbit(int x){
return x&(-x);
}
void add(int x,int d){
while(x < maxn){
sum[x] += d;
x += lowbit(x);
}
}
int getS(int x){
int ret = 0;
while(x > 0){
ret += sum[x];
x -= lowbit(x);
}
return ret;
}
int main(){
while(~scanf("%d%d",&n,&k)){
mma.clear();
memset(sum,0,sizeof sum);
for(int i = 1; i <= n; i++){
scanf("%d",&num[i]);
tn[i] = num[i];
}
sort(tn+1,tn+n+1,cmp);
int i = 1,cnt = 1;
while(i <= n){
mma[tn[i]] = cnt;
int j = i+1;
while(j <= n && tn[i]==tn[j]){
j++;
}
cnt++;
i = j;
}
long long ret = 0;
for(int i = 1; i <= n; i++){
ret += getS(mma[num[i]]-1);
add(mma[num[i]],1);
}
long long ans = ret-k;
if(ans < 0) ans = 0;
cout<<ans<<endl;
}
return 0;
}
HDU4911-Inversion(树状数组),布布扣,bubuko.com