逆序数 UVALive 6508 Permutation Graphs

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逆序数 UVALive 6508 Permutation Graphs

题目传送门 /* 题意:给了两行的数字,相同的数字连线,问中间交点的个数逆序数:第一行保存每个数字的位置,第二行保存该数字在第一行的位置,接下来就是对它求逆序数用归并排序或线段树求.想到了就简单了:) */ #include <cstdio> #include <algorithm> #include <cstring> #include <cmath> #include <vector> using namespace std; typed…

UVALive 6508 Permutation Graphs

Permutation Graphs Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit Status Practice UVALive 6508 #include<stdio.h> #include<string.h> ],a[],b[],c[],b1[]; long long num; void merg_sort(int a[],int l,int r) { int…

HDU 1394 Minimum Inversion Number（最小逆序数/暴力线段树树状数组归并排序）

题目链接: 传送门 Minimum Inversion Number Time Limit: 1000MS Memory Limit: 32768 K Description The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj. For a given sequence of numbe…

HDU-Minimum Inversion Number（最小逆序数）

Problem Description The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj. For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of…

逆序数2 HDOJ 1394 Minimum Inversion Number

题目传送门 /* 求逆序数的四种方法 */ /* 1. O(n^2) 暴力+递推法:如果求出第一种情况的逆序列,其他的可以通过递推来搞出来,一开始是t[1],t[2],t[3]....t[N] 它的逆序列个数是N个,如果把t[1]放到t[N]后面,逆序列个数会减少t[1]个,相应会增加N-(t[1]+1)个 */ #include <cstdio> #include <cstring> #include <algorithm> using namespace std;…

HDU 1394 Minimum Inversion Number（线段树/树状数组求逆序数）

Minimum Inversion Number Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 17737 Accepted Submission(s): 10763 Problem Description The inversion number of a given number sequence a1, a2, ...,…

Codeforces Gym 100463A Crossings 逆序数

Crossings Time Limit: 20 Sec Memory Limit: 256 MB 题目连接 http://codeforces.com/gym/100463 Description Given a permutation P of {0, 1, ..., n − 1}, we define the crossing number of it as follows. Write the sequence 0, 1, 2, . . . , n − 1 from left to ri…