2014 12 Program Design contest a title team

Input

Enter an integer you (n<=100000) and M, representing M's team have n players, have M operations (m<100000)

Next n-1 line, enter two numbers a and B, indicating A and B are good friends of friendship;

Next m line, there are the following operations (team number 1--n):

D x y means x and y are broken

Q x Y q indicates if X is a friend of the pot friends of the Y basin. Friends of the pot.

Output

For q operation, X is the pot friend of Y's .... Output Yes, otherwise output No.

Sample Input

6 5

1 3

3 5

1 2

3 4

2 6

D 1 3

Q 1 6

Q 1 4

D 3 5

Q 1 4

Sample output

Calculate yourself

Problem Solving Ideas:

When the game feels like my method will time out, I don't do it.

Code implementation (not submitted, do not know whether to timeout):

#include <stdio.h>intcont[100000][100000]={0};//record the relationship between the players//Cont[i][j] If 1, indicating that I,j is a good friend, for 1 means i,j is not a good friendintTEAMN,OPN;//number of players, number of operationsvoidDintXinty) {Cont[x][y]=0; CONT[Y][X]=0;}intFlag;voidQ (intXinty) {    if(x==y) {Flag=1; return; }    inti;  for(i=1; i<=teamn;i++)        if(Cont[i][x]) {cont[i][x]=0; Cont[x][i]=0;            Q (I,y); CONT[I][X]=1; Cont[x][i]=1; }}intMain () {scanf ("%d%d",&teamn,&OpN); inti; intTEAM1,TEAM2;//operation, Team 1, Team 2    CharOpra;  for(i=1; i<teamn;i++) {//record a good friendscanf"%d%d",&team1,&team2); CONT[TEAM1][TEAM2]=1; CONT[TEAM2][TEAM1]=1;    } getchar ();  for(i=0; i<opn;i++) {scanf ("%c%d%d",&opra,&team1,&team2); //printf ("%c\n", OPRA);GetChar (); //printf ("%c\n", OPRA);flag=0; if(opra=='D') D (TEAM1,TEAM2); Else if(opra=='Q'){        //printf ("\ \");Q (TEAM1,TEAM2); if(flag) printf ("yes\n"); Elseprintf"no\n"); }    //printf ("%d\n", flag);    }    return 0;}

Test data:

2014 12 Program Design contest a title team