HDU - 1394 Minimum Inversion Number【逆序数+dp】
2017-09-30 21:18
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Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 21660 Accepted Submission(s): 12952
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 the seqence, we will obtain another sequence. There are totally n such sequences as the following:
a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)
You are asked to write a program to find the minimum inversion number out of the above sequences.
Input
The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.
Output
For each case, output the minimum inversion number on a single line.
Sample Input
10
1 3 6 9 0 8 5 7 4 2
Sample Output
16
题意:
求一个数列的逆序数。不断把当前数列的第一个元素移到最后,构成新数列。求所有数列的最小逆序数。
题解:
可以发现中间的元素不会改变, 由于n个数是0~n-1,那么将第一个数a[0]移到最后时,可以很容易的求出逆序数的改变量。假如a[0]=3,那么后面就有3个比它小的数,他们都要-1,即最终-3;移到末尾后,其前面有6个比它大的数,此时这一位的逆序数为6。那么就可以推出转移方程了:
dp[i]=dp[i-1]-a[i-1]+n-a[i-1]-1;
代码:
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; #define inf 1<<29 int n,a[5005],dp[5005]; int cal() { int cnt=0; for(int i=1;i<n;i++) { for(int j=0;j<i;j++) if(a[j]>a[i]) cnt++; } return cnt; } int main() { while(~scanf("%d",&n)) { for(int i=0;i<n;i++) scanf("%d",&a[i]); memset(dp,0,sizeof dp); dp[0]=cal(); int minn=dp[0]; for(int i=1;i<n;i++) { dp[i]=dp[i-1]-a[i-1]+n-a[i-1]-1; minn=min(minn,dp[i]); } printf("%d\n",minn); } return 0; }
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