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  • hdu多校第3场C. Dynamic Graph Matching

    Problem C. Dynamic Graph Matching
    
    Time Limit: 8000/4000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
    Total Submission(s): 1215    Accepted Submission(s): 505
    
    
    Problem Description
    In the mathematical discipline of graph theory, a matching in a graph is a set of edges without common vertices.
    You are given an undirected graph with n vertices, labeled by 1,2,...,n. Initially the graph has no edges.
    There are 2 kinds of operations :
    + u v, add an edge (u,v) into the graph, multiple edges between same pair of vertices are allowed.
    - u v, remove an edge (u,v), it is guaranteed that there are at least one such edge in the graph.
    Your task is to compute the number of matchings with exactly k edges after each operation for k=1,2,3,...,n2. Note that multiple edges between same pair of vertices are considered different.
     
    
    Input
    The first line of the input contains an integer T(1≤T≤10), denoting the number of test cases.
    In each test case, there are 2 integers n,m(2≤n≤10,nmod2=0,1≤m≤30000), denoting the number of vertices and operations.
    For the next m lines, each line describes an operation, and it is guaranteed that 1≤u<v≤n.
     
    
    Output
    For each operation, print a single line containing n2 integers, denoting the answer for k=1,2,3,...,n2. Since the answer may be very large, please print the answer modulo 109+7.
     
    
    Sample Input
    1        
    4 8    
    + 1 2
    + 3 4
    + 1 3
    + 2 4
    - 1 2
    - 3 4
    + 1 2
    + 3 4
     
    
    Sample Output
    1 0
    2 1
    3 1
    4 2
    3 1
    2 1
    3 1
    4 2
     
    
    Source
    2018 Multi-University Training Contest 3
     
    
    Recommend
    chendu
     
    
    Statistic | Submit | Discuss | Note

    装压dp

    +:操作很简单就是:dp[i]+=dp[i-(1<<u)-(1<<v)];

    -:操作就想象成反的: dp[i]-=dp[i-(1<<u)-(1<<v)];拿总的减去用到用到u,v

    类似于背包某个物品不能放;

    #include <cstdio>
    #include <cstring>
    #include <iostream>
    //#include <algorithm>
    #include <vector>
    using namespace std;
    #define ll long long
    //#define mod 998244353
    const int mod=1e9+7;
    
    ll dp[1<<11];//dp[i]:i集合内点完全匹配的方案数
    int bit[1<<11];//i的二进制的1个数;
    ll ans[11];
    int lowbit(int x) {return x&-x;}
    int calc(int x)
    {
        int res=0;
        while(x){res++;x=x-lowbit(x);}
        return res;
    }
    int main()
    {
        for(int i=0;i<(1<<10);i++)
            bit[i]=calc(i);
        int T,n,m,u,v;
        char op[3];
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d",&n,&m);
            memset(dp,0,sizeof dp);
            memset(ans,0,sizeof ans);
            dp[0]=1;
            while(m--)
            {
                scanf("%s%d%d",op,&u,&v);
                u--;v--;
                if(op[0]=='+')
                {
                    for(int i=(1<<n)-1;i>0;i--)
                    if(((1<<u)&i)&&((1<<v)&i))
                    {
                        dp[i]+=dp[i-(1<<u)-(1<<v)];
                        ans[bit[i]]+=dp[i-(1<<u)-(1<<v)]; //对于i集合所有点的匹配,会对ans造成的影响
                        dp[i]=dp[i]%mod;
                        ans[bit[i]]=ans[bit[i]]%mod;
                    }
                }
                
                else
                {
                    for(int i=1;i<1<<n;i++)
                    if(((1<<u)&i)&&((1<<v)&i))
                    {
                        dp[i]-=dp[i-(1<<u)-(1<<v)];
                        ans[bit[i]]-=dp[i-(1<<u)-(1<<v)];
                        dp[i]=(dp[i]+mod)%mod;
                        ans[bit[i]]=(ans[bit[i]]+mod)%mod;
                    }
                }
                for(int i=2;i<=n;i=i+2)
                {
                    if(i!=2)
                        cout<<" ";
                    printf("%lld",ans[i]);
                    
                    
                }
                cout<<endl;
                
            }
            
            
            
            
        }
        
        
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/2014slx/p/9397606.html
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