pku1148 Utopia Divided

题目链接：http://acm.pku.edu.cn/JudgeOnline/problem?id=1148

题意简述：给定n，给定2*n个正整数，用这些数去描述点，然后给定几个数表示坐标轴象限，要求每次选取一个点(可以对给定的整数设置正负）使得与前面一个点的连线组成的向量所对应的坐标在这个象限内，当然把起始点设在(0,0)。求这写点。

解题思路：看到这题感觉没想法，看了一份英文版的解题报告，讲的很详细。再此就引用那一份解题报告了。

问题描述 – utopia.doc

• 问题分析

•         两个彼此独立的序列

•         对于一维问题求解

•         符号序列 si

•         数值序列 xi

•         对xi重新排列并加相应的正负号，使其按新顺序逐一相加后，等到符号si.

• Definition – alternating sequence

 

A sequence of non-zero integers X=(xa, xa+1,…, xb,), a≤b is an alternating sequence if

       1) |xa|< |xa+1|< …<|xb|, and

       2) for all i, a<i≤b, the sign of xi is different from the sign of xi-1.

Here, |xa| is the absolute value of xa.

• Lemma 1. Let X=(xa, xa+1,…, xb) be an alternating sequence. The sign of xb is equal to the sign of ∑a≤i≤bxi , the total sum of elements in X.

• Proof

Assume:

•         xb is positive

•         number of elements in X is even

• Then:
• xa+xa+1 , xa+2+xa+3 , … , xb-1+xb are all positive, thus the total sum ∑a≤i≤bxi is positive.

• Theorem 1.

Let X=(xa, xa+1,…, xb), a≤b be an alternating sequence, and let S=(sa, sa+1,…, sb) , a≤b be a sequence of signs. If the sign of xb is equal to sb , then there exists a sequence X’=(xia, xia+1,…, xib) such that

1)      {xa, xa+1,…, xb} ={xia, xia+1,…, xib}, and

2)      X’ is valid with respect to S.

3)      Proof

The proof is by induction on the number of k of elements in X. When k=1, it is easy to see that X’=X is a valid sequence with respect to S. Now we assume that k≥2. We let S1=S-sb, that is, S1=(sa,sa+1,…,sb-1).

Case 1. The sign of sb-1 is equal to xb ,

Let X1=X-xa, that is, X1=(xa+1,xa+2,…,xb).

Case 2. The sign of sb-1 is equal to xb-1 ,

Let X1=X-xb, that is, X1=(xa,xa+1,…,xb-1).

• Algorithm of utopia

 

Step 1. //read input

       1.1 read N;

       1.2 read 2N code numbers and            partition them into A and B such that |A|=|B|;

       1.3 read a sequence of regions     R=(r1, r2, …,rN );

Step 2. //find x-coordinates of code pairs

 

2.1 find a sequence of signs S=(s1, s2, …,sN ) such that for all j , 1≤j≤N, sj=‘+’ if rj=1,4; otherwise sj=‘-’.

2.2 find an alternating sequence X=(x1, x2, …,xN ) from A such that the sign of xN is equal to sN .

2.3 given X and S , find a valid sequence X’=(xi1, xi2, …,xiN ) w.r.t S according to the proof of Theorem 1.

Step 3. //find y-coordinates of code pairs

      

3.1 find a sequence of signs S=(s1, s2, …,sN ) such that for all j , 1≤j≤N, sj=‘+’ if rj=1,2; otherwise sj=‘-’.

3.2 find an alternating sequence Y=(y1, y2, …,yN ) from B such that the sign of yN is equal to sN .

3.3 given Y and S , find a valid sequence Y’=(yi1, yi2, …,yiN ) w.r.t S according to the proof of Theorem 1.

Step 4. //write output

       print (xi1,yi1),(xi2,yi2),…,(xiN,yiN).

代码：

#include<stdio.h>#include<algorithm>#include<iostream>using namespace std;int A[10001],B[10001],P[10001];char sx[10001],sy[10001];int N;void design(int D[],char s[],int l,int r,int n){ int i,j,k; if(n==1) { D[n]=P[l]; return; } if(s[n-1]==s[n]) { D[n]=P[l]; design(D,s,l+1,r,n-1); } else { D[n]=P[r]; design(D,s,l,r-1,n-1); }}int main(){ int i,j,k; scanf("%d",&N); for(i=1;i<=N;i++) scanf("%d",&A[i]); for(i=1;i<=N;i++) scanf("%d",&B[i]); for(i=1;i<=N;i++) scanf("%d",&P[i]); sort(A+1,A+1+N); sort(B+1,B+1+N); for(i=1;i<=N;i++) { if(P[i]==1||P[i]==4) sx[i]='+'; else sx[i]='-'; if(P[i]==1||P[i]==2) sy[i]='+'; else sy[i]='-'; } if(sx[N]=='-') { for(i=N;i>=1;i=i-2) A[i]=-A[i]; } else { for(i=N-1;i>=1;i=i-2) A[i]=-A[i]; } if(sy[N]=='-') { for(i=N;i>=1;i=i-2) B[i]=-B[i]; } else { for(i=N-1;i>=1;i=i-2) B[i]=-B[i]; } for(i=1;i<=N;i++) P[i]=A[i]; design(A,sx,1,N,N); for(i=1;i<=N;i++) P[i]=B[i]; design(B,sy,1,N,N); for(i=1;i<=N;i++) printf("%+d %+d/n",A[i],B[i]); return 0;}