題目
Constraints
Time Limit: 1 secs, Memory Limit: 32 MB
Description
N couples are standing in a circle, numbered consecutively clockwise from 1 to 2N. Husband and wife do not always stand together. We remove the couples who stand together until the circle is empty or we can't remove a couple any more.
Can we remove all the couples out of the circle?
Input
There may be several test cases in the input file. In each case, the first line is an integer N(1 <= N <= 100000)----the number of couples. In the following N lines, each line contains two integers ---- the numbers of each couple.
N = 0 indicates the end of the input.
Output
Output "Yes" if we can remove all the couples out of the circle. Otherwise, output "No".
Sample Input
4
1 4
2 3
5 6
7 8
2
1 3
2 4
0
Sample Output
Yes
No
思路
用數(shù)組模擬堆棧齿税，如果檢查到是一堆夫妻在一起，就將這對夫妻出棧，直到把所有夫妻都操作過一遍。最后判斷棧頂是否為空即可疏日。
代碼
#include<stdio.h>
#include<string.h>
int main() {
int i, j, n, a, b, c[200000 + 20], d[200000 + 20];
while (scanf("%d", &n) && n) {
memset(c, 0, sizeof(c));
memset(d, 0, sizeof(d));
for (i = 0; i < n; i++) {
scanf("%d %d", &a, &b);
c[a - 1] = 2 * i + 1;
c[b - 1] = 2 * i + 2;
}
for (i = 0, j = 0, d[0] = c[0]; i < 2 * n; i++) {
if (c[i + 1] != d[j] + 1) {
d[j + 1] = c[i + 1];
j++;
} else {
d[j] = 0;
j--;
}
}
printf(!d[0] ? "Yes\n" : "No\n");
}
return 0;
}
