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

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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

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Probability/Expected Value

Others

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Hardest

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

 bool Find(int x)
 {
     for (int i=; i<=n; i++)
     {
         if (!vis[i]&&Map[x][i]==)
         {
             vis[i]=;
             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=;
         memset(vis,,sizeof(vis));
         memset(Map,,sizeof(Map));
         memset(ok,,sizeof(ok));
         memset(ss,,sizeof(ss));
         for (int i=; i<=n; i++)
         {
             scanf("%d",&a[i]);
         }
         for (int i=; i<=m; i++)
         {
             scanf("%d",&b[i]);
             ss[b[i]]=;
         }
         for (int i=; i<=n; i++)
         {
             for (int j=i+; j<=n; j++)
             {
                 if (ss[a[i]+a[j]]==)
                     Map[i][j]=Map[j][i]=;
             }
         }
         for (int i=; i<=n; i++)
         {
             memset(vis,,sizeof(vis));
             if (Find(i))
                 ans++;
         }
         printf ("%d\n",ans/);
     }
     return ;
 }