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  • hdu5637 Transform (bfs+预处理)

    Problem Description
    A list of n integers are given. For an integer x you can do the following operations:

    + let the binary representation of x be b31b30...b0¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯, you can flip one of the bits.
    + let y be an integer in the list, you can change x to xy, where  means bitwise exclusive or operation.

    There are several integer pairs (S,T). For each pair, you need to answer the minimum operations needed to change S to T.
     

    Input
    There are multiple test cases. The first line of input contains an integer T (T20), indicating the number of test cases. For each test case:

    The first line contains two integer n and m (1n15,1m105) -- the number of integers given and the number of queries. The next line contains n integers a1,a2,...,an (1ai105), separated by a space.

    In the next m lines, each contains two integers si and ti (1si,ti105), denoting a query.
     

    Output
    For each test cases, output an integer S=(i=1mizi) mod (109+7), where zi is the answer for i-th query.
     

    Sample Input
    1 3 3 1 2 3 3 4 1 2 3 9
     

    Sample Output

    10

    题意:给你n个数,有m个询问,每一个询问有两个值a,b,每一次操作,你可以把a中的二进制的一位异或1,即那位从0变为1或者1变为0,或者你可以异或上n个数中的一个数,问最小变化的次数。一开始打算m个询问直接模拟,但是发现时间爆了,所以采用预处理的方案,然后每次询问花O(1)的时间。

    #include<iostream>
    #include<stdio.h>
    #include<stdlib.h>
    #include<string.h>
    #include<math.h>
    #include<vector>
    #include<map>
    #include<set>
    #include<queue>
    #include<stack>
    #include<string>
    #include<bitset>
    #include<algorithm>
    using namespace std;
    typedef long long ll;
    typedef long double ldb;
    #define inf 99999999
    #define pi acos(-1.0)
    #define maxn 1000050
    #define MOD 1000000007
    int vis[1<<(20)],a[20],dis[1<<(20)];
    int q[1111111][2];
    int n;
    int maxx;
    
    
    void bfs()
    {
        int i,j;
        memset(vis,0,sizeof(vis));
        int front,rear,x,y,xx,yy,state1;
        front=rear=1;
        q[rear][0]=0;q[rear][1]=0;
        vis[0]=1;dis[0]=0;
        while(front<=rear)
        {
            int state=q[front][0];
            int num=q[front][1];
            front++;
            for(i=0;i<18;i++){
                state1=(state^(1<<i));
                if(vis[state1 ]==0  ){
                    dis[state1 ]=num+1;
                    vis[state1 ]=1;
                    if(state1>200050)continue;
                    rear++;
                    q[rear][0]=state1;
                    q[rear][1]=num+1;
                }
            }
            for(i=1;i<=n;i++){
                state1=(state^a[i]);
                if(vis[state1 ]==0 ){
                    dis[state1 ]=num+1;
                    vis[state1 ]=1;
                    if(state1>200005)continue;
                    rear++;
                    q[rear][0]=state1;
                    q[rear][1]=num+1;
                }
            }
        }
    }
    
    
    
    int main()
    {
        int m,i,j,T;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d",&n,&m);
            int maxx=0;
            for(i=1;i<=n;i++){
                scanf("%d",&a[i]);
            }
            bfs();
            ll sum=0;
            int c,d;
            for(i=1;i<=m;i++){
                scanf("%d%d",&c,&d);
                sum=(sum+(ll)i*(ll)dis[c^d] )%MOD;
            }
            printf("%lld
    ",sum);
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/herumw/p/9464517.html
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