树形\(dp\)多半是跑不过去的，可用矩阵快速幂解决
题目大意:给定一个无向图，\(0\)秒时机器人在\(1\)号点，每秒可以走到另一个相邻的点或者不动或者自爆，求\(t\)秒内行动方案数
矩阵快速幂
分析:首先把问题统一，不动我们连自环即可，自爆可以连单向边到虚拟点，虚拟点连自环，然后问题就变成了\(t\)秒后走到每个点的方案数之和
假设用\(f[t][u]\)表示在\(t\)秒时走到\(u\)点的方案数，显然\(f[t][u] = \sum f[t-1][v]\)，有边\((u,v)\)，然后假设我们知道当前的\(f\)，把它构成一个矩阵
\(\begin{pmatrix}f(1),f(2)\dots f(n)\end{pmatrix}\)
然后我们发现将其乘上邻接矩阵\(G\)就可以得到下一秒的\(f\)值，然后就可以矩阵快速幂解决
#include <cstdio>
#include <cctype>
#include <cstring>
#include <vector>
using namespace std;
const int maxn = 32,mod = 2017;
inline int read(){
int x = 0;char c = getchar();
while(!isdigit(c))c = getchar();
while(isdigit(c))x = x * 10 + c - '0',c = getchar();
return x;
}
struct matrix{
int val[maxn][maxn],x,y;
matrix operator * (const matrix &rhs)const{
matrix res;
res.x = x;
res.y = rhs.y;
for(int i = 1;i <= res.x;i++)
for(int j = 1;j <= res.y;j++)res.val[i][j] = 0;
for(int i = 1;i <= res.x;i++)
for(int j = 1;j <= res.y;j++)
for(int k = 1;k <= y;k++)
res.val[i][j] = (res.val[i][j] + val[i][k] * rhs.val[k][j]) % mod;
return res;
}
}G,ans;
int n,m,t,out;
inline matrix qpow(const matrix &a,int b){
matrix res,base = a;
res.x = res.y = a.x;
for(int i = 1;i <= res.x;i++)
for(int j = 1;j <= res.y;j++)
res.val[i][j] = 0;
for(int i = 1;i <= res.x;i++)
res.val[i][i] = 1;
while(b){
if(b & 1)res = res * base;
base = base * base;
b >>= 1;
}
return res;
}
int main(){
n = read() + 1,m = read();
G.x = G.y = n;
for(int u,v,i = 1;i <= m;i++)
u = read(),v = read(),G.val[u][v] = G.val[v][u] = 1;
t = read();
for(int i = 1;i <= n;i++)G.val[i][i] = 1;
for(int i = 1;i <= n - 1;i++)G.val[i][n] = 1;
ans.x = 1,ans.y = n;
ans.val[1][1] = 1;
ans = ans * qpow(G,t);
for(int i = 1;i <= n;i++)
out = (out + ans.val[1][i]) % mod;
printf("%d\n",out);
return 0;
}