hdu 2157 How many ways?? (矩阵快速幂经典题目)

Posted by 111qqz on Wednesday, October 19, 2016

TOC

题意:给定一个有向图,问从A点恰好走k步(允许重复经过边)到达B点的方案数mod p的值

思路:
** 把给定的图转为邻接矩阵,即A(i,j)=1当且仅当存在一条边i->j。令C=A*A,那么C(i,j)=ΣA(i,k)A(k,j),实际上就等于从点i到点j恰好经过2条边的路径数(枚举k为中转点)。类似地,CA的第i行第j列就表示从i到j经过3条边的路径数。同理,如果要求经过k步的路径数,我们只需要快速幂求出A^k即可。**

M67_十个利用矩阵乘法解决的经典题目

这个转化好美啊。。。

/* ***********************************************
Author :111qqz
Created Time :Wed 19 Oct 2016 08:14:56 PM CST
File Name :code/hdu/2157.cpp
************************************************ */

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <cmath>
#include <cstdlib>
#include <ctime>
#define fst first
#define sec second
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ms(a,x) memset(a,x,sizeof(a))
typedef long long LL;
#define pi pair < int ,int >
#define MP make_pair

using namespace std;
const double eps = 1E-8;
const int dx4[4]={1,0,0,-1};
const int dy4[4]={0,-1,1,0};
const int inf = 0x3f3f3f3f;
const int N=25;
const int mod = 1000;
int n,m;
struct Mat
{
    int mat[N][N];
    void clear()
    {
    ms(mat,0);
    }
}M;

Mat operator * (Mat a,Mat b)
{
    Mat ans;
    ans.clear();
    for ( int i = 0 ; i < n ; i++)
    for ( int j = 0 ; j < n ;j++)
        for ( int k = 0 ; k < n ; k++)
        ans.mat[i][j] = (ans.mat[i][j] + a.mat[i][k] * b.mat[k][j])%mod;

    return ans;
}

Mat operator ^ (Mat a,int b)
{
    Mat ans;
    ans.clear();
    for ( int i = 0 ; i < n ; i++) ans.mat[i][i] = 1;

    while (b>0)
    {
    if (b&1) ans = ans * a;
    b = b >> 1LL;
    a = a* a;
    }
    return ans;
}
LL ksm( LL a,LL b, LL k)
{
    LL res = 1LL;
    while (b)
    {
    if (b&1) res = (res * a) % k;
    b = b >> 1LL;
    a = ( a * a) % k;
    }
    return res;
}
int main()
{
    #ifndef  ONLINE_JUDGE 
    freopen("code/in.txt","r",stdin);
  #endif

    while (~scanf("%d%d",&n,&m))
    {
        if (n==0&&m==0) break;
        M.clear();
        for ( int i = 1 ; i <= m ; i ++)
        {
        int s,t;
        scanf("%d %d",&s,&t);
        M.mat[s][t] = 1;
        }

        int T;
        scanf("%d",&T);
        while (T--)
        {
        int a,b,k;
        scanf("%d%d%d",&a,&b,&k);
        Mat ans;
        ans.clear();
        ans = M ^ k;
        printf("%d\n",ans.mat[a][b]);
        }
    }

  #ifndef ONLINE_JUDGE  
  fclose(stdin);
  #endif
    return 0;
}

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