转移矩阵+快速幂HDU5607
题意：http://acm.hdu.edu.cn/showproblem.php?pid=5607
有向图中问你从定点u走到x的概率
思路：
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\\Users\\13606\\Desktop\\Input.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr strcat 13 #include <string> 14 #include <time.h>// srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 #include <cassert> 21 #include <iomanip> 22 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 23 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 24 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 25 //****************** 26 clock_t __START,__END; 27 double __TOTALTIME; 28 void _MS(){__START=clock();} 29 void _ME(){__END=clock();__TOTALTIME=(double)(__END-__START)/CLOCKS_PER_SEC;cout<<"Time: "<<__TOTALTIME<<" s"<<endl;} 30 //*********************** 31 #define rint register int 32 #define fo(a,b,c) for(rint a=b;a<=c;++a) 33 #define fr(a,b,c) for(rint a=b;a>=c;--a) 34 #define mem(a,b) memset(a,b,sizeof(a)) 35 #define pr printf 36 #define sc scanf 37 #define ls rt<<1 38 #define rs rt<<1|1 39 typedef pair<int,int> PII; 40 typedef vector<int> VI; 41 typedef unsigned long long ull; 42 typedef long long ll; 43 typedef double db; 44 const db E=2.718281828; 45 const db PI=acos(-1.0); 46 const ll INF=(1LL<<60); 47 const int inf=(1<<30); 48 const db ESP=1e-9; 49 const int mod=(int)1e9+7; 50 const int N=60; 51 52 struct Mat 53 { 54 ll mat[N][N]; 55 Mat operator*(const Mat &a)const 56 { 57 Mat b=Mat(); 58 for(int i=1;i<=55;++i) 59 { 60 for(int j=1;j<=55;++j) 61 { 62 for(int k=1;k<=55;++k) 63 { 64 b.mat[i][j]=(b.mat[i][j]+mat[i][k]*a.mat[k][j]%mod); 65 b.mat[i][j]%=mod; 66 } 67 } 68 } 69 return b; 70 } 71 }start; 72 Mat Mqpow(long long n,int x) 73 { 74 Mat ans=Mat(); 75 for(int i=1;i<=x;++i) 76 ans.mat[i][i]=1; 77 while(n) 78 { 79 if(n&1) 80 ans=ans*start; 81 start=start*start; 82 n>>=1; 83 } 84 return ans; 85 } 86 ll qpow(ll a,ll b,ll mod) 87 { 88 ll ans; 89 // a%=mod; 90 ans=1; 91 while(b!=0) 92 { 93 if(b&1) 94 ans=(a*ans)%mod;//注意左乘右乘; 95 b/=2; 96 a=(a*a)%mod; 97 } 98 return ans; 99 } 100 ll cnt[N]; 101 102 int main() 103 { 104 int n,m; 105 while(~sc("%d%d",&n,&m)) 106 { 107 mem(cnt,0); 108 start=Mat(); 109 for(int i=1;i<=m;++i) 110 { 111 int u,v; 112 sc("%d%d",&u,&v); 113 start.mat[u][v]++; 114 cnt[u]++; 115 } 116 for(int i=1;i<=n;++i) 117 for(int j=1;j<=n;++j) 118 if(start.mat[i][j]) 119 start.mat[i][j]=start.mat[i][j]*qpow(cnt[i],mod-2,mod)%mod; 120 int ask; 121 sc("%d",&ask); 122 Mat temp=start; 123 for(int i=1;i<=ask;++i) 124 { 125 int u,k; 126 sc("%d%d",&u,&k); 127 start=temp; 128 Mat ans=Mqpow(k,n); 129 for(int j=1;j<=n;++j) 130 pr("%lld ",ans.mat[u][j]); 131 pr("\n"); 132 } 133 } 134 return 0; 135 } 136 137 /**************************************************************************************/
