zoj 2362 Beloved Sons【二分匹配】

题目：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2361

来源：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=26760#problem/B

Beloved Sons

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Time Limit: 2 Seconds      Memory Limit: 65536 KB      Special Judge

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Once upon a time there lived a king and he had N sons. And the king wanted to marry his beloved sons on the girls that they did love. So one day the king asked his sons to come to his room and tell him whom do they love.

But the sons of the king were all young men so they could not tell exactly whom they did love. Instead of that they just told him the names of the girls that seemed beautiful to them, but since they were all different, their choices of beautiful girls also did not match exactly.

The king was wise. He did write down the information that the children have provided him with and called you, his main wizard.

"I want all my kids to be happy, you know," he told you, "but since it might be impossible, I want at least some of them to marry the girl they like. So please, prepare the marriage list."

"I want all my kids to be happy, you know," he told you, "but since it might be impossible, I want at least some of them to marry the girl they like. So please, prepare the marriage list."

So, go on, make a list to maximize the king's happiness.

This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

Input

The first line of the input file contains N - the number of king's sons (1 <= N <= 400). The second line contains N integer numbers A_iranging from 1 to 1000 - the measures of king's love to each of his sons.

Next N lines contain lists of king's sons' preferences - first K_i - the number of the girls the i-th son of the king likes, and then K_i integer numbers - the girls he likes (all potentially beautiful girls in the kingdom were numbered from 1 to N, you know, beautiful girls were rare in those days).

Output

Output N numbers - for each son output the number of the beautiful girl he must marry or 0 if he must not marry the girl he likes.

Denote the set of sons that marry a girl they like by L, then you must maximize the value of

Sample Input

1

4
1 3 2 4
4 1 2 3 4
2 1 4
2 1 4
2 1 4

Sample Output

2 1 0 4

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Author: Andrew Stankevich
Source: Andrew Stankevich's Contest #3

题意：

就是一个匹配问题。

国王有N个儿子，有 N 个漂亮的女生和他们匹配。

但是不能够满足每个儿子都娶到中意的姑娘，所以要最大限度的让最少的儿子失望。

首先输入多组测试数据。T

然后对于个测试数据，输入 N 表示有几个儿子。

下面一行有 N 个数，表示国王对相应编号儿子的喜爱程度。

【这里有个猫腻，国王肯定是最宠爱他最喜欢的儿子了，所以会先满足最喜欢的儿子】

剩下 N 行，相应代表第一个儿子到第 N个儿子的喜好

每一行第一个数据表示喜换的姑娘的个数 num，后面 num 个数表示喜欢的姑娘的编号

最后让你输出第一个儿子到最后一个儿子娶到的姑娘的编号

注意：王子们对他们喜欢的姑娘们的喜欢程度都是相同的，如果没有娶到喜欢的姑娘则不娶，输出 0

算法：二分匹配

思路：

不能完全套用原来的二分匹配的模板，自己要改一下。

首先要把王子们按照国王对他们的喜欢程度排序【从大到小排序，最喜欢的先选，注意记录王子的编号】

然后建图就是一般的简单的建图，map[i][j] = 0 表示王子 i 不喜欢姑娘 j ; 相应的 map[i][j] = 1就表示喜欢了没有什么好说的了。

再就是按照排序好的顺序依次匹配了。

最后输出的时候注意：match[i] = index; 表示的是姑娘匹配的王子，而题目要求的是输出王子匹配的姑娘编号，自己调整下就好了。

原有的二分匹配模板：【每一个点都要匹配完】

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int maxn = 410;int map[maxn][maxn];int match[maxn];bool vis[maxn];int uN, vN;bool dfs(int u){    for(int v = 1; v <= vN; v++)    {        if(!vis[v] && map[u][v])        {            vis[v] = true;            if(match[v] == -1 || dfs(match[v]))            {                match[v] = u;                return true;            }        }    }   return false;}bool hungary(){    int sum = 0;    memset(match, -1, sizeof(match));    for(int j = 1 ; j <= uN; j++)    {        memset(vis, false, sizeof(vis));        if(dfs(i)) sum++;    }    if(sum == uN) return true;    else return false;}

推荐题目：

POJ 1469 COURSES【匈牙利算法入门 二分图的最大匹配 模板题】

按照题目调整的模板：

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int maxn = 410;int map[maxn][maxn];int match[maxn];bool vis[maxn];int uN, vN;struct Son{    int index;    int a;}son[maxn];bool dfs(int u){    for(int v = 1; v <= vN; v++)    {        if(!vis[v] && map[u][v])        {            vis[v] = true;            if(match[v] == -1 || dfs(match[v]))            {                match[v] = u;                return true;            }        }    }   return false;}void hungary(){    int sum = 0;    memset(match, -1, sizeof(match));    for(int j = 1 ; j <= uN; j++) //按照国王喜爱程度匹配    {        int i = son[j].index;        memset(vis, false, sizeof(vis));        dfs(i);    }}

BAccepted844 KB260 msC++ (g++ 4.4.5)

#include<stdio.h>#include<string.h>#include<algorithm>using namespace std;const int maxn = 410;int map[maxn][maxn];int match[maxn];bool vis[maxn];int uN, vN;struct Son{    int index; //编号    int a; //国王喜爱程度}son[maxn];int ans[maxn];bool cmp(Son A, Son B){    return A.a >= B.a;}bool dfs(int u){    for(int v = 1; v <= vN; v++)    {        if(!vis[v] && map[u][v])        {            vis[v] = true;            if(match[v] == -1 || dfs(match[v]))            {                match[v] = u;                return true;            }        }    }   return false;}void hungary(){    int sum = 0;    memset(match, -1, sizeof(match));    for(int j = 1 ; j <= uN; j++) //按照国王的喜爱程度依次匹配    {        int i = son[j].index;        memset(vis, false, sizeof(vis));        dfs(i);    }}int main(){    int n;    int T;    scanf("%d", &T);    while(T--)    {        scanf("%d", &n);        uN = vN = n;        memset(map, 0, sizeof(map));        for(int i = 1; i <= n; i++)        {            scanf("%d", &son[i].a);            son[i].index = i;        }        sort(son+1,son+(n+1),cmp); // 按照喜爱程度排序        int num;        for(int i = 1; i <= n; i++) //简单建图        {            scanf("%d", &num);            int index;            for(int j = 1; j <= num; j++)            {                scanf("%d", &index);                map[i][index] = 1;            }        }        hungary();        memset(ans,0,sizeof(ans));        for(int i = 1; i <= n; i++) //调整结果 map[i] 表示的是第 i 个姑娘匹配的王子编号        {            if(match[i] != -1) ans[match[i]] = i;        }        for(int i = 1; i <= n; i++)        {            if(i == 1) printf("%d", ans[i]);            else printf(" %d", ans[i]);        }        printf("\n");    }    return 0;}