SOJ-4074（强联通分量的需要加几条边）

 

/****************************************************************************************************** 
 ** Copyright (C) 2011.07.01-2013.07.01 
 ** Author: famousDT <[email protected]> 
 ** Edit date: 2011-10-03 
******************************************************************************************************/  
#include <stdio.h>   
#include <stdlib.h>//abs,atof(string to float),atoi,atol,atoll   
#include <math.h>//atan,acos,asin,atan2(a,b)(a/b atan),ceil,floor,cos,exp(x)(e^x),fabs,log(for E),log10   
#include <vector>   
#include <queue>    
#include <map>   
#include <time.h>   
#include <set>   
#include <stack>   
#include <string>   
#include <iostream>   
#include <assert.h>   
#include <string.h>//memcpy(to,from,count   
#include <ctype.h>//character process:isalpha,isdigit,islower,tolower,isblank,iscntrl,isprll   
#include <algorithm>   
using namespace std;  
  
//typedef long long ll;   
  
#define MY_PI acos(-1)   
#define MY_MAX(a, b) ((a) > (b) ? (a) : (b))      
#define MY_MIN(a, b) ((a) < (b) ? (a) : (b))   
#define MY_MALLOC(n, type) ((type *)malloc((n) * sizeof(type)))   
#define MY_ABS(a) (((a) >= 0) ? (a) : (-(a)))   
#define MY_INT_MAX 0x7fffffff   
  
/*==========================================================*\ 
| max（出度为0的点的个数，入度为0的点的个数）就是此题的答案
| http://hi.baidu.com/c%C3%D42/blog/item/f5efe0a844749daaca130c32.html
\*==========================================================*/  

struct my
{
	int a, b;
}wo[100 + 1];

int main()  
{  
    int m, n;
	int i;
	int x, y;
	while (scanf("%d%d", &m, &n) == 2) {
		for (i = 0; i <= m; ++i)
			wo[i].a = wo[i].b = 0;
		for (i = 0; i < n; ++i) {
			scanf("%d%d", &x, &y);
			wo[x].b++;
			wo[y].a++;
		}
		int maxa = 0, maxb = 0;
		for (i = 1; i <= m; ++i) {
			if (wo[i].a == 0)
				maxa++;
			if (wo[i].b == 0)
				++maxb;
		}
		printf("%d\n", MY_MAX(maxa, maxb));
	}
    return 0;  
}