@Dmaxiya
2022-12-04T22:24:41.000000Z
字数 6618
阅读 222
出题
当 时,由于数据范围很小,完全可以暴力深搜计算到达每个节点的方案数,最后将与节点 相邻的方案数相加取余就是答案,时间复杂度为 ,这种做法可以通过这部分数据。
对于统计图中路径方案数,可以先创建一个邻接矩阵 M[i][j]
(下标从 开始),M[i][j]=M[j][i]=x
表示节点 与节点 之间有 条无向边,注意当 时,M[u][v]
只能被计算一次;由于大禹不能走回家里,只能有以 为起点的边,不能有以 为终点的边,所以 G[i][1]
需要手动清零。
两个邻接矩阵相乘,得到的第 行第 列的值,就是从节点 出发,走两步能到达节点 的方案数,题目要求 步以内的方案数总和,即求以下公式的结果:
当 时,由于 非常大,每次矩阵乘法的时间复杂度都为 ,若使用顺序乘法,总共需要计算 次,必然超时,需要优化。
考虑计算 ,有以下递推式:
当 时, 的数据范围达到 ,上一种解法已不适用,观察到节点 与膜数 的值非常小,考虑计算循环节,由于矩阵 第一列元素全为 ,之后无论如何相乘第一列都必然为 ,因此可以忽略第一列的状态变化。
除去第 列共有 个元素在不断变化,可知循环节长度必然小于等于 ,因此最多只要计算出前 项 即可。
循环节状态必然为:
由于 非常大,但 对 取余前后再对 做减法的结果是一致的,因此可以不必做大数减法计算,先用 对 取余后再对 做减法同样可以得到正确的循环节下标。
时间复杂度为 ,最坏情况下需要执行约 次计算。
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 100;
int T, n, m, k, p, ans, u, v;
int cnt[maxn];
bool isDoorway[maxn];
vector<int> G[maxn];
void dfs(int root, int k) {
++cnt[root];
if (k == 0) {
return ;
}
for (auto pos: G[root]) {
if (pos == 1) {
continue;
}
dfs(pos, k - 1);
}
}
int main() {
cin >> T;
cin >> n >> m >> k >> p;
for (int i = 0; i < m; ++i) {
cin >> u >> v;
if (u > v) {
swap(u, v);
}
G[u].push_back(v);
if (u == 1) {
isDoorway[v] = true;
} else {
if (u != v) {
G[v].push_back(u);
}
}
}
dfs(1, k);
for (int i = 2; i <= n; ++i) {
if (isDoorway[i]) {
ans += cnt[i];
}
}
cout << ans % p << endl;
return 0;
}
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int matrixSize = 50;
const int maxn = 100000 + 100;
int MOD;
bool isDoorway[50];
LL add(LL a, LL b) {
a += b;
if(a >= MOD) {
return a - MOD;
}
return a;
}
struct Matrix {
int size;
LL num[matrixSize][matrixSize];
const Matrix& operator=(const Matrix &m) {
size = m.size;
for(int i = 0; i < size; ++i) {
memcpy(num[i], m.num[i], sizeof(LL) * size);
}
return (*this);
}
void init() {
for(int i = 0; i < size; ++i) {
memset(num[i], 0, sizeof(LL) * size);
num[i][i] = 1;
}
}
void clear() {
for (int i = 0; i < size; ++i) {
memset(num[i], 0, sizeof(LL) * size);
}
}
void set(int n, int m) {
size = n;
int u, v;
for (int i = 0; i < m; ++i) {
scanf("%d%d", &u, &v);
--u;
--v;
++num[u][v];
if (u != v) {
++num[v][u];
}
}
for (int i = 0; i < size; ++i) {
for (int j = 0; j < size; ++j) {
if (j == 0) {
num[i][j] = 0;
} else {
if (i == 0) {
isDoorway[j] = (num[i][j] != 0);
}
num[i][j] %= MOD;
}
}
}
}
const Matrix& operator*(const Matrix &m) const {
static Matrix ans;
ans.size = m.size;
for(int i = 0; i < size; ++i) {
for(int j = 0; j < size; ++j) {
ans.num[i][j] = 0;
for(int k = 0; k < size; ++k) {
ans.num[i][j] = add(ans.num[i][j], num[i][k] * m.num[k][j] % MOD);
}
}
}
return ans;
}
const Matrix& operator*=(const Matrix &m) {
(*this) = (*this) * m;
return *this;
}
const Matrix& operator+(const Matrix &m) const {
static Matrix ans;
ans.size = m.size;
for (int i = 0; i < size; ++i) {
for (int j = 0; j < size; ++j) {
ans.num[i][j] = add(num[i][j], m.num[i][j]);
}
}
return ans;
}
const Matrix& operator+=(const Matrix &m) {
for (int i = 0; i < size; ++i) {
for (int j = 0; j < size; ++j) {
num[i][j] = add(num[i][j], m.num[i][j]);
}
}
return *this;
}
LL hash() {
LL ret = 0;
for (int i = 0; i < size; ++i) {
for (int j = 1; j < size; ++j) {
ret = ret * 10 + num[i][j];
}
}
return ret;
}
};
Matrix one, zero;
struct MMatrix {
int size;
Matrix m[2][2];
void set(const Matrix &m) {
size = 2;
this->m[0][0] = m;
this->m[0][1] = one;
this->m[1][0] = zero;
this->m[1][1] = one;
}
const MMatrix& operator=(const MMatrix &mm) {
size = mm.size;
for(int i = 0; i < size; ++i) {
for (int j = 0; j < size; ++j) {
m[i][j] = mm.m[i][j];
}
}
return (*this);
}
void init() {
for(int i = 0; i < size; ++i) {
m[i][i] = one;
m[i][i ^ 1] = zero;
}
}
const MMatrix& operator*=(const MMatrix &mm) {
static MMatrix ans;
ans.size = size;
for(int i = 0; i < size; ++i) {
for(int j = 0; j < size; ++j) {
ans.m[i][j] = zero;
for(int k = 0; k < size; ++k) {
ans.m[i][j] += m[i][k] * mm.m[k][j];
}
}
}
(*this) = ans;
return ans;
}
void fastPow(LL n) {
static MMatrix ans;
ans.size = size;
for(ans.init(); n != 0; n >>= 1) {
if((n & 1) == 1) {
ans *= (*this);
}
(*this) *= (*this);
}
(*this) = ans;
}
};
int T, n, m, k, ans;
Matrix matrix, matrixSum;
MMatrix mmatrix;
char kStr[maxn];
LL h;
LL pow10[20];
map<LL, int> hashIndex;
vector<LL> hs;
int getNum(LL h, int x, int y) {
int r = n - 1 - x;
int c = n - 1 - y;
return h / pow10[r * (n - 1) + c] % 10;
}
void solve1() {
scanf("%d%d%d%d", &n, &m, &k, &MOD);
matrix.set(n, m);
one.size = n;
one.init();
zero.size = n;
zero.clear();
mmatrix.set(matrix);
mmatrix.fastPow(k);
mmatrix.m[0][1] *= matrix;
ans = 0;
for (int i = 1; i < n; ++i) {
if (isDoorway[i]) {
ans = (ans + mmatrix.m[0][1].num[0][i]) % MOD;
}
}
printf("%d\n", ans);
}
void solve2() {
pow10[0] = 1;
for (int i = 1; i < 18; ++i) {
pow10[i] = pow10[i - 1] * 10;
}
scanf("%d%d%s%d", &n, &m, kStr, &MOD);
matrix.set(n, m);
matrixSum.size = n;
matrixSum.clear();
h = matrixSum.hash();
do {
hashIndex[h] = hs.size();
hs.push_back(h);
matrixSum = matrixSum * matrix + matrix;
h = matrixSum.hash();
} while (hashIndex.find(h) == hashIndex.end());
int loopIndex = hashIndex[h];
int loopLength = hs.size() - loopIndex;
int index = 0;
for (int i = 0; kStr[i] != '\0'; ++i) {
index = (index * 10 + (kStr[i] - '0')) % loopLength;
}
index = ((index - loopIndex) % loopLength + loopLength) % loopLength + loopIndex;
for (int j = 1; j < n; ++j) {
if (isDoorway[j]) {
ans = (ans + getNum(hs[index], 0, j)) % MOD;
}
}
printf("%d\n", ans);
}
int main() {
scanf("%d", &T);
if (T <= 30) {
solve1();
return 0;
}
solve2();
return 0;
}
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int testCases = 50;
ofstream fout;
int T, n, m, k, p, u, v;
string numToStr(int x) {
string str;
while (x != 0) {
str = char('0' + x % 10) + str;
x /= 10;
}
return str;
}
string getFileName(int index) {
return "data/" + numToStr(index) + ".in";
}
int getRand(int l, int r) {
return (LL)rand() * rand() * rand() % (r - l + 1) + l;
}
void writeToFile() {
for (int i = 0; i < m; ++i) {
u = getRand(1, n);
v = getRand(1, n);
fout << u << " " << v << endl;
}
}
int main() {
srand(time(0));
// data in 20%, 2 <= n,m,k <= 7, p = 1e9+7
int one = testCases * 20 / 100;
for (int i = 1; i <= one; ++i) {
fout.open(getFileName(i));
fout << i << endl;
n = getRand(5, 7);
m = getRand(5, 7);
k = getRand(5, 7);
p = 1000000007;
fout << n << " " << m << " " << k << " " << p << endl;
writeToFile();
fout.close();
}
// data in 60%, 2 <= n <= 50, 1 <= m <= 1e6, 1 <= k <= 1e6, 1e7 <= p <= 1e9
int two = testCases * 60 / 100;
for (int i = one + 1; i <= two; ++i) {
fout.open(getFileName(i));
fout << i << endl;
n = getRand(45, 50);
m = getRand(100000, 1000000);
k = getRand(999900, 1000000);
p = getRand(10000000, 1000000000);
fout << n << " " << m << " " << k << " " << p << endl;
writeToFile();
fout.close();
}
// data in 100%, 2 <= n <= 4, 1 <= m <= 1e6, 1 <= k <= 1e100000, 1 <= p <= 3
for (int i = two + 1; i <= testCases; ++i) {
fout.open(getFileName(i));
fout << i << endl;
n = getRand(3, 4);
m = getRand(100000, 1000000);
p = getRand(2, 3);
int lenOfk = getRand(50000, 100000);
fout << n << " " << m << " ";
fout << getRand(1, 9);
for (int j = 1; j < lenOfk; ++j) {
fout << getRand(0, 9);
}
fout << " " << p << endl;
writeToFile();
fout.close();
}
return 0;
}