UVALive 3989 Ladies' Choice

Ladies' Choice

Time Limit: 6000ms

Memory Limit: 131072KB

This problem will be judged on  UVALive. Original ID: 3989
64-bit integer IO format: %lld      Java class name: Main

Background

Teenagers from the local high school have asked you to help them with the organization of next years Prom. The idea is to find a suitable date for everyone in the class in a fair and civilized way. So, they have organized a web site where all students, boys and girls, state their preferences among the class members, by ordering all the possible candidates. Your mission is to keep everyone as happy as possible. Assume that there are equal numbers of boys and girls.

Problem

Given a set of preferences, set up the blind dates such that there are no other two people of opposite sex who would both rather have each other than their current partners. Since it was decided that the Prom was Ladies' Choice, we want to produce the best possible choice for the girls.

 

Input

Input consists of multiple test cases the first line of the input contains the number of test cases. There is a blank line before each dataset. The input for each dataset consists of a positive integer N, not greater than 1,000, indicating the number of couples in the class. Next, there are N lines, each one containing the all the integers from 1 to N, ordered according to the girls preferences. Next, there are N lines, each one containing all the integers from 1 to N, ordered according to the boys preferences.

 

Output

The output for each dataset consists of a sequence of N lines, where the i-th line contains the number of the boy assigned to the i-th girl (from 1 to N). Print a blank line between datasets.

 

Sample Input

1

5

1 2 3 5 4

5 2 4 3 1

3 5 1 2 4

3 4 2 1 5

4 5 1 2 3

2 5 4 1 3

3 2 4 1 5

1 2 4 3 5

4 1 2 5 3

5 3 2 4 1

Sample Output

1

2

5

3

4

Source

Regionals 2007, Europe - Southwestern

 

解题：稳定婚姻问题

 

男士按照各位女士在自己心目中的地位从高到底，依次求婚。

 

如果女士先前没有对象，被求婚，那么暂且凑一对，

如果女士先前有对象，如果此时求婚的男士比她原来的那个对象在她心目中的地位更高，那么与原来的那个男士解除关系，原来的男士变成光棍。

与新求婚的那位男士建立对象关系。如果此时求婚的男士没有女士原先的男士在她心中地位高，则此时的男士继续光棍着。

 

直到没有光棍为止，此时的婚姻关系稳定。

 

 1 #include <bits/stdc++.h>

 2 using namespace std;

 3 const int maxn = 1010;

 4 int n,mr[maxn][maxn],miss[maxn][maxn];

 5 int wife[maxn],husband[maxn],nxt[maxn];

 6 queue<int>q;

 7 void engage(int man,int woman) {

 8     if(husband[woman]) {

 9         wife[husband[woman]] = 0;

10         q.push(husband[woman]);

11     }

12     husband[woman] = man;

13     wife[man] = woman;

14 }

15 void GaleShapley() {

16     while(!q.empty()) {

17         int man = q.front();

18         q.pop();

19         int woman = mr[man][nxt[man]++];

20         if(!husband[woman]) engage(man,woman);

21         else if(miss[woman][husband[woman]] > miss[woman][man])

22             engage(man,woman);

23         else q.push(man);

24     }

25 }

26 int main() {

27     int kase,tmp;

28     scanf("%d",&kase);

29     while(kase--) {

30         scanf("%d",&n);

31         while(!q.empty()) q.pop();

32         for(int i = 1; i <= n; ++i) {

33             for(int j = 1; j <= n; ++j)

34                 scanf("%d",mr[i]+j);

35             nxt[i] = 1;

36             wife[i] = 0;

37             q.push(i);

38         }

39         for(int i = 1; i <= n; ++i) {

40             for(int j = 1; j <= n; ++j) {

41                 scanf("%d",&tmp);

42                 miss[i][tmp] = j;

43             }

44             husband[i] = 0;

45         }

46         GaleShapley();

47         for(int i = 1; i <= n; ++i)

48             printf("%d\n",wife[i]);

49         if(kase) putchar('\n');

50     }

51     return 0;

52 }

View Code