LightOJ 1184 - Marriage Media 【二分图最大匹配】

来源：互联网发布：双色球绝密算法编辑：程序博客网时间：2024/05/16 14:13

题目链接：http://www.lightoj.com/volume_showproblem.php?problem=1184

根据一些要求建图即可

代码：

#include <iostream>  #include <algorithm>  #include <set>  #include <map>  #include <string.h>  #include <queue>  #include <sstream>  #include <stdio.h>  #include <math.h>  #include <stdlib.h>  using namespace std;int n, m;int p[1000][1000];int book[1000];int match[1000];int dfs(int u){    int i;    for (i = 1; i <= m; i++)    {        if (book[i] == 0 && p[u][i] == 1)        {            book[i] = 1;            if (match[i] == 0 || dfs(match[i]))            {                match[i] = u;                return 1;            }        }    }    return 0;}struct peo{    int hig;    int age;    int ok;}a[110], b[110];int main(){    int t;    int cases = 1;    scanf("%d", &t);    while (t--)    {        scanf("%d%d", &n, &m);        for (int i = 1; i <= n; i++) scanf("%d%d%d", &a[i].hig, &a[i].age, &a[i].ok);        for (int i = 1; i <= m; i++) scanf("%d%d%d", &b[i].hig, &b[i].age, &b[i].ok);        int ans = 0;        memset(match, 0, sizeof(match));        memset(p, 0, sizeof(p));        for (int i = 1; i <= n; i++)            for (int j = 1; j <= m; j++)            {                if (a[i].ok == b[j].ok && abs(a[i].hig - b[j].hig) <= 12 && abs(a[i].age - b[j].age) <= 5)                p[i][j] = 1;            }        for (int i = 1; i <= n; i++)        {            memset(book, 0, sizeof(book));            if (dfs(i))                ans++;        }        printf("Case %d: %d\n", cases++, ans);    }    return 0;}

0 0