半年不碰 C++ 实力明显变弱
个人贡献三题,零罚时比较让人欣慰,至少手还算稳
(还是太弱了
# One car comes and one car goes
听说是小学奥数题,没啥好说的
# Color
裸的树形 DP ,穷举一下每个点选各种 type 的种类数就好,剩下的就是乘法原理吧
#include <bits/stdc++.h> | |
#define MOD 10000009 | |
#define N 50000 | |
#define M 30 | |
using namespace std; | |
int n, m, s, t; | |
bool can[N][M]; | |
vector<int> a[N]; | |
long long f[N][M]; | |
inline long long nis(int c, int i) { | |
static int ret; | |
ret = f[c][0] - f[c][i]; | |
return ret < 0 ? ret + MOD : ret; | |
} | |
void dfs(int t, int p) { | |
int len = a[t].size(), c; | |
for (int i = 0; i < len; i++) | |
if (a[t][i] != p) { | |
dfs(a[t][i], t); | |
} | |
for (int i = 1; i <= m; i++) { | |
if (can[t][i]) | |
f[t][i] = 1; | |
else continue ; | |
for (int j = 0; j < len; j++) { | |
if ((c = a[t][j]) == p) | |
continue ; | |
f[t][i] *= nis(c, i); | |
f[t][i] %= MOD; | |
} | |
} | |
for (int i = 1; i <= m; i++) { | |
f[t][0] += f[t][i]; | |
f[t][0] %= MOD; | |
} | |
} | |
int main() { | |
srand(time(0)); | |
ios::sync_with_stdio(0); | |
while (cin >> n >> m) { | |
memset(f, 0, sizeof(f)); | |
for (int i = 1; i <= n; i++) | |
a[i].clear(); | |
for (int i = 1; i < n; i++) { | |
cin >> s >> t; | |
a[s].push_back(t); | |
a[t].push_back(s); | |
} | |
for (int i = 1; i <= n; i++) | |
for (int j = 1; j <= m; j++) { | |
cin >> t; | |
can[i][j] = t; | |
} | |
dfs(s = rand() % n + 1, 0); | |
cout << f[s][0] << endl; | |
} | |
return 0; | |
} |
# Lost in WHU
给一个无向图,问在 T 步之内有多少条路线从 1 走到 n
结论题,直接建立图的邻接矩阵,然后求矩阵 T 次幂就好
由于题目要求到了 n 就不能走出去,所以需要删掉 n 所有出边,然后 n 自环一下就好
但是由于太弱并不知道还有自环这种操作,一股脑删光了所有射出边,给邻接矩阵多加了一层,用一种十分不优雅的方式侥幸过了 XD
#include <bits/stdc++.h> | |
#define N 100 | |
#define MOD 1000000007LL | |
using namespace std; | |
typedef long long LL; | |
typedef vector<vector<LL>> mat; | |
int n, m; | |
inline void init(mat &x, bool I) { | |
x.resize(n + 1); | |
for (int i = 0; i <= n; i++) | |
x[i].resize(n + 1); | |
for (int i = 0; i <= n; i++) | |
for (int j = 0; j <= n; j++) | |
x[i][j] = 0; | |
if (I) for (int i = 0; i <= n; i++) | |
x[i][i] = 1; | |
} | |
inline mat mul(const mat &a, const mat &b) { | |
mat c; init(c, false); | |
for (int i = 0; i <= n; i++) | |
for (int j = 0; j <= n; j++) { | |
for (int k = 0; k <= n; k++) { | |
c[i][j] += (a[i][k] * b[k][j]) % MOD; | |
c[i][j] %= MOD; | |
} | |
} | |
return c; | |
} | |
inline mat quick_mod(mat a, int q) { | |
mat c; init(c, true); | |
while (q) { | |
if (q & 1) | |
c = mul(a, c); | |
a = mul(a, a); | |
q >>= 1; | |
} | |
return c; | |
} | |
inline void print(const mat &a) { | |
for (int i = 0; i <= n; i++, cout << endl) | |
for (int j = 0; j <= n; j++) | |
cout << a[i][j] << " "; | |
} | |
int main() { | |
int x, y, t; | |
cin >> n >> m; | |
mat a; init(a, false); | |
while (m--) { | |
cin >> x >> y; | |
x--, y--; | |
a[x][y] = 1; | |
a[y][x] = 1; | |
} | |
cin >> t; | |
for (int i = 0; i < n; i++) | |
a[n - 1][i] = 0; | |
a[n - 1][n] = 1; | |
a[n][n] = 1; | |
a = quick_mod(a, t); | |
a[0][n] += a[0][n - 1]; | |
cout << a[0][n] % MOD << endl; | |
return 0; | |
} |