Codeforces 514E Darth Vader and Tree DP + 矩阵快速幂

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题目大意:

给定n和x, (n <= 10^5, x <= 10^9) 表示现在有一颗完全n叉树, 根节点为起始点, 然后给定n个di分别表示每个对应的分支的距离, (1 <= di <= 100), 然后对于所有的点, 求其中到根节点的距离不超过x的节点的个数, 由于数字可能很大, 对最后的结果模上10^9 + 7之后输出

大致思路：

思路写在代码注释里了

代码如下：

Result : Accepted Memory : 852 KB Time : 1045 ms

/* * Author: Gatevin * Created Time:  2015/2/25 11:49:15 * File Name: poi~.cpp */#include<iostream>#include<sstream>#include<fstream>#include<vector>#include<list>#include<deque>#include<queue>#include<stack>#include<map>#include<set>#include<bitset>#include<algorithm>#include<cstdio>#include<cstdlib>#include<cstring>#include<cctype>#include<cmath>#include<ctime>#include<iomanip>using namespace std;const double eps(1e-8);typedef long long lint;/* * 首先注意到每个di <= 100数据很小, 如果用ti表示所有n中分支中长度为i的个数, 那么就有t[1~100],  * 用dp[i]表示到根节点长度为i的点的个数, 那么不难发现状态转移方程 * dp[x] = dp[x - 1]*t[1] + dp[x - 2]*t[2] + ... + dp[x - 100]*t[100] * 显然对于这个线性的状态转移方程, 预先处理计算出dp[0~99]即可递推后边的值 * 最后的结果是S[x] = dp[0] + dp[1] + ... + dp[x] * 所以可以建立矩阵转移, 由于 * {dp[x], dp[x - 1], ... , dp[x - 99], S[x]} * |  t[1]  1 0 ... 0 t[1]  | *                                              |  t[2]  0 1 ... 0 t[2]  | *                                              |............... 1 t[3]  | *                                              | t[100] 0 0 ... 0 t[100]| *                                              |    0   0 0 ... 0   1   | * = {dp[x + 1], dp[x], dp[x - 1], ... dp[x - 98], S[x + 1]} * 所以对于转移矩阵求快速幂即可, 很普通的矩阵快速幂, 想到dp的转移方程就很容易想到了 */const lint mod = 1000000007LL;int n, x, t[110];struct Matrix{    lint a[110][110];    Matrix()    {        memset(a, 0, sizeof(a));        for(int i = 1; i <= 101; i++)            a[i][i] = 1;    }};Matrix operator + (Matrix m1, Matrix m2){    Matrix ret;    for(int i = 1; i <= 101; i++)        for(int j = 1; j <= 101; j++)            ret.a[i][j] = (m1.a[i][j] + m2.a[i][j]) % mod;    return ret;}Matrix operator * (Matrix m1, Matrix m2){    Matrix ret;    for(int i = 1; i <= 101; i++)        for(int j = 1; j <= 101; j++)        {            ret.a[i][j] = 0;            for(int k = 1; k <= 101; k++)                ret.a[i][j] = (ret.a[i][j] + m1.a[i][k]*m2.a[k][j] % mod) % mod;        }    return ret;}Matrix quick_pow(Matrix base, int pow){    Matrix I;    while(pow)    {        if(pow & 1)            I = I*base;        base = base*base;        pow >>= 1;    }    return I;}lint dp[110];int main(){    scanf("%d %d", &n, &x);    int tt;    memset(t, 0, sizeof(t));    for(int i = 1; i <= n; i++)    {        scanf("%d", &tt);        t[tt]++;    }    Matrix A;    memset(A.a, 0, sizeof(A.a));    for(int i = 1; i <= 100; i++)        A.a[i][1] = A.a[i][101] = t[i];    for(int i = 1; i < 100; i++)        A.a[i][i + 1] = 1;    A.a[101][101] = 1;    memset(dp, 0, sizeof(dp));    dp[0] = 1;    for(int i = 1; i <= 100; i++)        for(int j = 1; i - j >= 0 && j <= 100; j++)            dp[i] = (dp[i - j]*t[j] + dp[i]) % mod;    if(x <= 100)    {        lint ans = 0;        for(int i = 0; i <= x; i++) ans = (ans + dp[i]) % mod;        cout<<ans<<endl;        return 0;    }    A = quick_pow(A, x - 99);    lint S = 0;    for(int i = 0; i <= 99; i++) S = (S + dp[i]) % mod;    lint ans = 0;    for(int i = 1; i <= 100; i++)        ans = (ans + dp[100 - i]*A.a[i][101]) % mod;    ans = (ans + S*A.a[101][101]) % mod;    cout<<ans<<endl;    return 0;}

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