ARC097E Sorted and Sorted

传送门

题目

There are 2N balls, N white and N black, arranged in a row. The integers from 1 through N are written on the white balls, one on each ball, and they are also written on the black balls, one on each ball. The integer written on the i-th ball from the left (1 ≤ i ≤ 2N) is a_i, and the color of this ball is represented by a letter c_i. c_i = W represents the ball is white; c_i = B represents the ball is black.

Takahashi the human wants to achieve the following objective:

• For every pair of integers (i,j) such that 1 ≤ i < j ≤ N, the white ball with i written on it is to the left of the white ball with j written on it.
• For every pair of integers (i,j) such that 1 ≤ i < j ≤ N, the black ball with i written on it is to the left of the black ball with j written on it.

In order to achieve this, he can perform the following operation:

• Swap two adjacent balls.

Find the minimum number of operations required to achieve the objective.

Constraints

• 1 ≤ N ≤ 2000
• 1 ≤ a_i ≤ N
• c_i = W or c_i = B.
• If i ≠ j, (a_i,c_i) ≠ (a_j,c_j)

Input

Input is given from Standard Input in the following format:

N
c1 a1
c2 a2
:
c2N a2N

Output

Print the minimum number of operations required to achieve the objective.

题目大意

给出一个黑棋N个白棋N个的排列，每一种颜色的球分别标上1 - N，每次可以交换相邻两个球，求白棋相对顺序正确并且黑棋相对顺序正确，所需要最少的步数

分析

设dpij为排好白球前i个和黑球前j个所需最小步数，用a和b数组分别记录黑白球的每一数值所在位置，用c1和c2分别记录将黑白球某一数值x移到此颜色的x-1数值之后所需步数（这里的步数既有同色数目又有异色数目）。所以我们不难想出dpij由dpi-1j和dpij-1转移而来，因此得到方程式dp[i][j]=min(dp[i-1][j]+c1[i][j],dp[i][j-1]+c2[i][j])。

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<queue>
#include<vector>
#include<set>
#include<map>
#include<stack>
using namespace std;
int dp[2100][2100],a[2100],b[2100],c1[2100][2100],c2[2100][2100];
int main(){
      int n,m,i,j,k,x;
      char ch;
      cin>>n;
      for(i=1;i<=2*n;i++){
           cin>>ch>>x;
           if(ch=='B'){
               a[x]=i;
           }else {
               b[x]=i;
           }
      }
      for(i=1;i<=n;i++){
           for(j=1;j<i;j++)
              if(a[i]<a[j])c1[i][0]++;
           for(j=1;j<=n;j++){
               c1[i][j]=c1[i][j-1];
               if(b[j]>a[i])c1[i][j]++;
           }
      }
      //c用于记录要向后走几步
      for(i=1;i<=n;i++){
           for(j=1;j<i;j++)
              if(b[i]<b[j])c2[0][i]++;
           for(j=1;j<=n;j++){
               c2[j][i]=c2[j-1][i];
               if(a[j]>b[i])c2[j][i]++;
           }
      }
      int wh1=0,wh2=0;
      for(i=1;i<=n;i++){
           wh1+=c1[i][0];
           wh2+=c2[0][i];
           dp[i][0]=wh1;
           dp[0][i]=wh2;
      }
      for(i=1;i<=n;i++)
         for(j=1;j<=n;j++)
            dp[i][j]=min(dp[i-1][j]+c1[i][j],dp[i][j-1]+c2[i][j]);
      cout<<dp[n][n]<<endl;
      return 0;
}