Continuation and collection of learning notes--gang Gang

A word does not first affixed to the topic description in a city live n person, any two people who know is not a friend is the enemy, and satisfies: 1, my friend's friend is my friend, 2, my enemy's enemy is my friend; all the people who are friends form a gang. Tell you about this n person's m piece of information, that is, some two people are friends, or some two people are enemies, ask you to write a program to calculate how many of the city may have a number of groups? Input 1th behavior N and m,n less than 1000,m less than 5000, the following M-line, each behavior p x y,p value is 0 or 1,p is 0 o'clock, indicating that X and Y are friends, p is 1 o'clock, that x and Y are enemies. Output An integer that indicates that n individuals may have a maximum of several gangs. Sample Input6
4
E 1 4
F 3 5
F 4 6
E 1 2
Sample Output3
HINT

{1},{2,4,6},{3,5}

The first idea, like the ABC beast, is to maintain the relationship between each point and ancestor. And it seems more simple. So we have the following procedure.

#include <iostream>#include<cstdio>using namespacestd;intfa[1010],deep[1010],ans[1010][2],n,m;intGETF (intk) {    if(fa[k]!=k) {        intt=Fa[k]; FA[K]=GETF (Fa[k]); DEEP[K]=deep[k]+Deep[t]; DEEP[K]=deep[k]%2; }    returnfa[k];}intMain () {CIN>>n>>m;  for(intI=1; i<=n;++i) {Fa[i]=i; Deep[i]=0; }     for(intI=1; i<=m;++i) {        Charc[2]; intx, y; scanf ("%s%d%d",&c,&x,&y); if((c[0]=='1') && (GETF (x)! =getf (y))) {            intfx=GETF (x); intfy=getf (y); DEEP[FX]= (deep[y]-deep[x]+3)%2; FA[FX]=fy; }        if((c[0]=='0') && (GETF (x)! =getf (y))) {            intfx=GETF (x); intfy=getf (y); DEEP[FX]= (deep[y]-deep[x]+2)%2; FA[FX]=fy; }    }         for(intI=1; i<=n;++i) {Fa[i]=GETF (i); //cout<<fa[i]<< "" <<deep[i]<<endl;ans[fa[i]][deep[i]]++; }    intAdd=0;  for(intI=1; i<=n;++i) for(intj=0; j<=1; ++j)if(ans[i][j]>0) {Add++; } cout<<add; return 0;}

And then 、、、、、、 blew 0.

Where's the problem? Suddenly found that the problem is not: My enemy's friend is my enemy. mdzz! It's not logical, though Noi never tells the logic.

Start again and think of the points of the way. Connected by a friend, Fa[x]=y, fa[x]=y+n,fa[x+n]=y by an enemy. The problem is solved perfectly. Downstairs program:

#include <cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespacestd; intn,m,ans,a[1010]; intfa[2020]; intFind (intx) {if(!fa[x]| | fa[x]==x)returnfa[x]=x; returnfa[x]=Find (fa[x]); }  voidUnion (intXinty) {x=find (x); y=Find (y); if(x==y)return ; FA[X]=y; }  intMain () {intI,x,y; Charp[Ten]; CIN>>n>>m;  for(i=1; i<=m;i++) {scanf ("%s%d%d",p,&x,&y); if(p[0]=='F') Union (x, y); ElseUnion (x, y+n), Union (x+n,y); }       for(i=1; i<=n;i++) A[i]=Find (i); Sort (a+1, a+n+1);  for(i=1; i<=n;i++)          if(i==1|| a[i]!=a[i-1])              ++ans; cout<<ans<<Endl; return 0; }  

To be continue ...

Continuation and collection of learning notes--gang Gang