归并排序求逆序对数：
和归并排序一样，划分和递归求解都好理解，关键在于合并，对于右边的j ，统计出左边比j 大 的元素个数 f(j)，所有的f(j)家和就是我们要的逆序对数！
在归并排序中，我们将右边的元素向临时数组中加入的时候，左边还没加入得便是比j 大的元素！ 既有m-p个
在加右边时，不断累加m-p即可！
以 POJ 1804 为例！
- #include <cstdio>
- #include <cstring>
- #include <algorithm>
-
- using namespace std;
- int a[5007],b[5007],n,sum,ks;
- void sort(int x,int y){
- if (x + 1 >= y)return;
- int m = x + (y-x)/2;
- sort(x,m); sort(m,y);
- int p = x, q = m, i = x;
- while (p < m || q < y){
- if (q >= y || (p < m && a[p] <= a[q])) b[i++] = a[p++];
- else b[i++] = a[q++], sum+=m-p;
- }
- for (int i = x; i < y; ++i) a[i] = b[i];
- }
- int main(){
- int T;
- scanf("%d",&T);
- while(T--){
- scanf("%d",&n);
- for (int i = 0; i < n; ++i)scanf("%d",a+i);
- sum = 0;
- sort(0,n);
- if (ks++)puts("");
- printf("Scenario #%d:\n%d\n",ks,sum);
- }
- return 0;
- }
| Brainman
Description Background Raymond Babbitt drives his brother Charlie mad. Recently Raymond counted 246 toothpicks spilled all over the floor in an instant just by glancing at them. And he can even count Poker cards. Charlie would love to be able to do cool things like that, too. He wants to beat his brother in a similar task. Problem Here's what Charlie thinks of. Imagine you get a sequence of N numbers. The goal is to move the numbers around so that at the end the sequence is ordered. The only operation allowed is to swap two adjacent numbers. Let us try an example: swap (2 8) 8 2 0 3 swap (2 0) 8 0 2 3 swap (2 3) 8 0 3 2 swap (8 0) 0 8 3 2 swap (8 3) 0 3 8 2 swap (8 2) 0 3 2 8 swap (3 2) 0 2 3 8 swap (3 8) 0 2 8 3 swap (8 3) 0 2 3 8 So the sequence (2 8 0 3) can be sorted with nine swaps of adjacent numbers. However, it is even possible to sort it with three such swaps: swap (8 0) 2 0 8 3 swap (2 0) 0 2 8 3 swap (8 3) 0 2 3 8 The question is: What is the minimum number of swaps of adjacent numbers to sort a given sequence?Since Charlie does not have Raymond's mental capabilities, he decides to cheat. Here is where you come into play. He asks you to write a computer program for him that answers the question. Rest assured he will pay a very good prize for it. Input The first line contains the number of scenarios. For every scenario, you are given a line containing first the length N (1 <= N <= 1000) of the sequence,followed by the N elements of the sequence (each element is an integer in [-1000000, 1000000]). All numbers in this line are separated by single blanks. Output Start the output for every scenario with a line containing "Scenario #i:", where i is the number of the scenario starting at 1. Then print a single line containing the minimal number of swaps of adjacent numbers that are necessary to sort the given sequence. Terminate the output for the scenario with a blank line. Sample Input 4 4 2 8 0 3 10 0 1 2 3 4 5 6 7 8 9 6 -42 23 6 28 -100 65537 5 0 0 0 0 0 Sample Output Scenario #1: 3 |
Scenario #2:
0
Scenario #3:
5
Scenario #4:
0
Source
TUD Programming Contest 2003, Darmstadt, Germany
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