1345 Legion training camp

Description

Mr. S is the commander of the Missile Corps. He has a group of soldiers capable of missile technology. Each soldier has a technical level score and is divided into one group of troops based on the technical level score. Several regiments form a temporary missile force. Today he is interested and wants to merge the two missile troops. However, he hopes that all the technology groups will be arranged in order to form a team from high to low, of course, after the same technology group is merged, the two teams are merged into one team.

Now he is too tired, because he has been practicing missile technology for too long yesterday ...... So I hope you can help him merge the team ......

The rules for providing army intelligence information are as follows:

For any group of two troops, their technical score is di (0 <= di <= 2000), and their team has AI (1 <= AI <= 10000) personnel.

For a single army, the information is

A1 d1

A2 D2

...

An DN

0 0

0 0 is used as the end mark. Two troops are provided.

I hope that you will return the information of the merged troops and deliver the information from low to high according to missile technology.

Input

The first line of input data is T, indicating that T groups of data exist;

The following T blocks contain information about two troops. The specific rules are described.

Output

For each group of data, information about the combined troops is output, sorted in ascending order by technical score.

 

Just sort questions by simulation.

#include <stdio.h>main(){int number,te;    double a1[2001],a2[2001];   double b1[2001],b2[2001];   double c1[4002],c2[4002];double temp1,temp2;int up1,up2,up3;int i,j;scanf("%d",&number);for(te=1;te<=number;te++){scanf("%lf %lf",&temp1,&temp2);up1=0;up2=0;while(!(temp1==0&&temp2==0)){a1[up1]=temp1;a2[up1]=temp2;up1++;scanf("%lf %lf",&temp1,&temp2);}       scanf("%lf %lf",&temp1,&temp2);    while(!(temp1==0&&temp2==0)){b1[up2]=temp1;b2[up2]=temp2;up2++;          scanf("%lf %lf",&temp1,&temp2);}up3=0;       for(i=0;i<up1;i++)   for(j=0;j<up2;j++)   {   if(a2[i]==b2[j]&&a1[i]!=-1)   { c1[up3]=a1[i]+b1[j];     c2[up3]=a2[i]; up3++; a1[i]=-1; b1[j]=-1;   }   }      for(i=0;i<up1;i++)  if(a1[i]!=-1)  {   c1[up3]=a1[i];      c2[up3]=a2[i];       up3++;  }for(j=0;j<up2;j++)if(b1[j]!=-1){c1[up3]=b1[j];c2[up3]=b2[j];up3++;}for(i=0;i<up3;i++)for(j=i+1;j<up3;j++){if(c2[i]>c2[j]){temp1=c2[j];temp2=c1[j];c2[j]=c2[i];c1[j]=c1[i];c2[i]=temp1;c1[i]=temp2;}}for(i=0;i<up3;i++)printf("%.0lf %.0lf\n",c1[i],c2[i]);}}