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 x⊕y, 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 (T≤20), indicating the number of test cases. For each test case:

The first line contains two integer n and m (1≤n≤15,1≤m≤105) -- the number of integers given and the number of queries. The next line contains n integers a1,a2,...,an (1≤ai≤105), separated by a space.

In the next m lines, each contains two integers si and ti (1≤si,ti≤105), denoting a query.

 

Output

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

 

Sample Input

13 31 2 33 41 23 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 1000000007int 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\n",sum);    }    return 0;}