uva 10763 Foreign Exchange-优快云博客

本文链接：https://blog.youkuaiyun.com/yanshiqin4/article/details/18595029

本文介绍了一种解决大规模学生外币兑换需求匹配的问题。通过整理和排序学生的需求信息，利用特定算法判断是否存在有效的匹配方案。适用于非营利组织管理大规模国际学生交换项目。

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Problem E
Foreign Exchange
Input: standard input
Output: standard output
Time Limit: 1 second

Your non-profit organization (iCORE - international Confederation of Revolver Enthusiasts) coordinates a very successful foreign student exchange program. Over the last few years, demand has sky-rocketed and now you need assistance with your task.

The program your organization runs works as follows: All candidates are asked for their original location and the location they would like to go to. The program works out only if every student has a suitable exchange partner. In other words, if a student wants to go from A to B, there must be another student who wants to go from B to A. This was an easy task when there were only about 50 candidates, however now there are up to 500000 candidates!

Input

The input file contains multiple cases. Each test case will consist of a line containing n - the number of candidates (1≤n≤500000), followed by n lines representing the exchange information for each candidate. Each of these lines will contain 2 integers, separated by a single space, representing the candidate's original location and the candidate's target location respectively. Locations will be represented by nonnegative integer numbers. You may assume that no candidate will have his or her original location being the same as his or her target location as this would fall into the domestic exchange program. The input is terminated by a case where n = 0; this case should not be processed.

Output

For each test case, print "YES" on a single line if there is a way for the exchange program to work out, otherwise print "NO".

Sample Input Output for Sample Input

1 2

2 1

3 4

4 3

100 200

200 100

57 2

2 57

1 2

2 1

1 2

3 4

5 6

7 8

9 10

11 12

13 14

15 16

17 18

19 20

YES

description:

整数对的匹配问题。

方法：

1.先对整数对做一下处理，比较数对中两个数（a， b)的大小，并把较大的数交换到前面(即a)，较小的数交换到后面（即b）标记是否做了交换。

2.排序。（先对数对中第一个数进行排序，若相等对第二个数进行排序）。

3.排序之后，相同的数对就聚集在一起。进行统计。

代码：

#include 
#include 
#define N 500000
typedef struct pat pat;
struct pat{
    int a;
    int b;
    int direct;
};
pat pat_arr[N], *p_pat_arr[N];
int comp(const void *x, const void *y)
{
    pat **l = (pat **)x;
    pat **u = (pat **)y;
    if((*l)->a != (*u)->a)  return (*l)->a - (*u)->a;
    else return (*l)->b - (*u)->b;
}
int judge(pat *p[], int n)
{
    int i, j, s;
    for(i = 0; i < n; i++)
        p[i] = pat_arr + i;
    qsort(p, n, sizeof(pat *), comp);
    i = 0;
    while(i < n){
        s = 0;
        for(j = i; j < n && p[j]->a == p[i]->a 
            && p[j]->b == p[i]->b; j++)
            s += p[j]->direct;
        if(s != 0)
            break;
        i = j;
    }
    return i == n;
}
int main()
{
    int i, j, n, x, y;
    while(scanf("%d", &n) != EOF && n != 0){
        for(i = 0; i < n; i++){
            scanf("%d %d", &x, &y);
            if(x > y){
                pat_arr[i].a = x;
                pat_arr[i].b = y;
                pat_arr[i].direct = 1;
            }
            else{
                pat_arr[i].a = y;
                pat_arr[i].b = x;
                pat_arr[i].direct = -1;
            }
        }
        printf("%s\n", judge(p_pat_arr, n) ? "YES" : "NO");
    }
    return 0;
}