BNUOJ 12756 Social Holidaying（二分匹配）

题目链接：http://www.bnuoj.com/bnuoj/problem_show.php?pid=12756

Social Holidaying

Time Limit: 3000ms

Memory Limit: 131072KB

 

This problem will be judged on UVALive. Original ID: 5874
64-bit integer IO format: %lld      Java class name: Main

Prev Submit Status Statistics Discuss Next

Font Size: + -

Type:   None   Graph Theory       2-SAT       Articulation/Bridge/Biconnected Component       Cycles/Topological Sorting/Strongly Connected Component       Shortest Path           Bellman Ford           Dijkstra/Floyd Warshall       Euler Trail/Circuit       Heavy-Light Decomposition       Minimum Spanning Tree       Stable Marriage Problem       Trees       Directed Minimum Spanning Tree       Flow/Matching           Graph Matching               Bipartite Matching               Hopcroft–Karp Bipartite Matching               Weighted Bipartite Matching/Hungarian Algorithm           Flow               Max Flow/Min Cut               Min Cost Max Flow   DFS-like       Backtracking with Pruning/Branch and Bound       Basic Recursion       IDA* Search       Parsing/Grammar       Breadth First Search/Depth First Search       Advanced Search Techniques           Binary Search/Bisection           Ternary Search   Geometry       Basic Geometry       Computational Geometry       Convex Hull       Pick's Theorem   Game Theory       Green Hackenbush/Colon Principle/Fusion Principle       Nim       Sprague-Grundy Number   Matrix       Gaussian Elimination       Matrix Exponentiation   Data Structures       Basic Data Structures       Binary Indexed Tree       Binary Search Tree       Hashing       Orthogonal Range Search       Range Minimum Query/Lowest Common Ancestor       Segment Tree/Interval Tree       Trie Tree       Sorting       Disjoint Set   String       Aho Corasick       Knuth-Morris-Pratt       Suffix Array/Suffix Tree   Math       Basic Math       Big Integer Arithmetic       Number Theory           Chinese Remainder Theorem           Extended Euclid           Inclusion/Exclusion           Modular Arithmetic       Combinatorics           Group Theory/Burnside's lemma           Counting       Probability/Expected Value   Others       Tricky       Hardest       Unusual       Brute Force       Implementation       Constructive Algorithms       Two Pointer       Bitmask       Beginner       Discrete Logarithm/Shank's Baby-step Giant-step Algorithm       Greedy       Divide and Conquer   Dynamic Programming                    Tag it!

Source

Regionals 2011, Asia - Kuala Lumpur

 

题目大意：先输入一个P表示有几组测试数据，再输入一个n，m分别表示A、B集合中分别有多少元素。在A集合中找到多少对数相加=B集合中的元素。A集合中的数不可以重复，但是B集合中的数字可以重复出现。

解题思路：二分匹配。先将A集合中的和的所有可能情况列出来，然后在B集合中搜索即可。

 

详见代码。

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

int ss[9999999],Map[410][410];
int ok[410],vis[410];
int a[410],b[410];
int n,m;

bool Find(int x)
{
    for (int i=1; i<=n; i++)
    {
        if (!vis[i]&&Map[x][i]==1)
        {
            vis[i]=1;
            if (!ok[i])
            {
                ok[i]=x;
                return true;
            }
            else
            {
                if (Find(ok[i]))
                {
                    ok[i]=x;
                    return true;
                }
            }
        }
    }
    return false;
}

int main()
{
    int t;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d",&n,&m);
        int ans=0;
        memset(vis,0,sizeof(vis));
        memset(Map,0,sizeof(Map));
        memset(ok,0,sizeof(ok));
        memset(ss,0,sizeof(ss));
        for (int i=1; i<=n; i++)
        {
            scanf("%d",&a[i]);
        }
        for (int i=1; i<=m; i++)
        {
            scanf("%d",&b[i]);
            ss[b[i]]=1;
        }
        for (int i=1; i<=n; i++)
        {
            for (int j=i+1; j<=n; j++)
            {
                if (ss[a[i]+a[j]]==1)
                    Map[i][j]=Map[j][i]=1;
            }
        }
        for (int i=1; i<=n; i++)
        {
            memset(vis,0,sizeof(vis));
            if (Find(i))
                ans++;
        }
        printf ("%d
",ans/2);
    }
    return 0;
}