patd乙级真题 1065. 单身狗(25)

题目链接

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思路

定义couple数组，couple[i] == j,若j == 100001代表i没对象，否则i的对象为j。定义people数组，people[i] == j，若j == 0，则i没来聚会，若j == 1，则i来了。

代码

#include <stdio.h>
#include <stdlib.h>
#include <ctype.h>
#include <string.h>

int main(){
	int couple_num,people_num,i,temp1_num,temp2_num;
	int couple[100012],     //couple[i] == j 代表i的对象是j
		people[100012]={0}, //people[i] == 0 代表i没参加聚会
		solo_num,temp_solo_num;
	for(i = 0;i<100012;i++){//couple[i] == 100001代表i没对象
		couple[i] = 100001;
	}

	char temp1[6],temp2[6];
	scanf("%d",&couple_num);
	for(i = 0;i< couple_num;i++){
		scanf("%s %s",temp1,temp2);
		temp1_num = atoi(temp1);
		temp2_num = atoi(temp2);
		couple[temp1_num] = temp2_num; 
		couple[temp2_num] = temp1_num;
	}
	scanf("%d",&people_num);
	for(i = 0;i<people_num;i++){ //记录到场的人
		scanf("%s",temp1);
		temp1_num =atoi(temp1);
		people[temp1_num] = 1;
	}
	for(i = 0,solo_num = 0;i<100000;i++){
		if(people[i] && couple[i] == 100001 || people[i]&&!people[couple[i]]) //人到了但没对象的或者人到了对象没到的都是单身狗
			solo_num++;
	}
	printf("%d\n",solo_num);
	
	for(i = 0,temp_solo_num = 0;i<100000;i++){
		if(people[i] && couple[i] == 100001 || people[i]&&!people[couple[i]]){
			
			if(temp_solo_num <solo_num -1)
				printf("%05d ",i);
			else
				printf("%05d",i);
			temp_solo_num++;
		}
	}
	return 0;
}