题目链接
A.Knight Moves
题面
思路
马的bfs、最短路径问题。模板题
示例代码
#include<bits/stdc++.h>using namespace std;#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;struct point{int x, y;point(int x, int y) : x(x), y(y) {}
};int dx[8] = {-2, -1, 1, 2, 2, 1, -1, -2};
int dy[8] = {1, 2, 2, 1, -1, -2, -2, -1};void dfs(string start, string end) {if(start == end) {cout << "To get from " << start << " to " << end << " takes " << 0 << " knight moves.\n";return;}queue<point> q;point start_point(start[0] - 'a', start[1] - '1'), end_point(end[0] - 'a', end[1] - '1');q.push(start_point);vector<vector<bool>> vis(8, vector<bool>(8, false));vis[start_point.x][start_point.y] = true;vector<vector<int>> dist(8, vector<int>(8, inf));dist[start_point.x][start_point.y] = 0;while(!q.empty()) {point curr = q.front();q.pop();if(curr.x == end_point.x && curr.y == end_point.y) {cout << "To get from " << start << " to " << end << " takes " << dist[curr.x][curr.y] << " knight moves.\n";return;}fer(i, 0, 8){int nx = curr.x + dx[i], ny = curr.y + dy[i];auto OnRange = [] (int a, int b) {return a >= 0 && a < 8 && b >= 0 && b < 8;};if(OnRange(nx, ny) && !vis[nx][ny]) {if(dist[curr.x][curr.y] + 1 < dist[nx][ny]) {dist[nx][ny] = dist[curr.x][curr.y] + 1;q.push({nx, ny});vis[nx][ny] = true;}}}}}signed main() {ios::sync_with_stdio(false); cin.tie(nullptr);string s1, s2;while(cin >> s1 >> s2) {dfs(s1, s2);}return 0;
}
B.Gold Transportation
题面
示例代码
#include <bits/stdc++.h>using namespace std;#define ll long long
// #define int ll
#define pii pair<int, int>
#define all(x) x.begin(), x.end()
#define fer(i, m, n) for (int i = m; i < n; ++i)
#define ferd(i, m, n) for (int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'const int inf = 0x3f3f3f3f;
const int N = 1001, M = 100001;int head[N], nex[M], to[M], val[M], now;
void add(int a, int b, int v)
{to[++now] = b;val[now] = v;nex[now] = head[a];head[a] = now;to[++now] = a;val[now] = 0;nex[now] = head[b];head[b] = now;
}//*********************int sp, ep, d[N];int bfs()
{queue<int> Q;memset(d, -1, sizeof(d));d[sp] = 0;Q.push(sp);while (!Q.empty()){int p = Q.front();Q.pop();for (int i = head[p]; ~i; i = nex[i]){int u = to[i];if (d[u] == -1 && val[i] > 0){d[u] = d[p] + 1;Q.push(u);}}}return d[ep] != -1;
}int dfs(int p, int v)
{int r = 0;if (p == ep)return v;for (int i = head[p]; (~i) && r < v; i = nex[i]){int u = to[i];if (val[i] > 0 && d[u] == d[p] + 1){int x = dfs(u, min(val[i], v - r));r += x;val[i] -= x;val[i ^ 1] += x;}}if (!r)d[p] = -2;return r;
}int dinic()
{int ans = 0, t;while (bfs()){while (t = dfs(sp, inf))ans += t;}return ans;
}//***********************void init()
{now = -1; // 要求第一条边为0memset(head, -1, sizeof(head));
}int n, m, sum;
int a[M], b[M], v[M];
int s[N], g[N];int judge(int limit)
{init();sp = 0, ep = 201;for (int i = 1; i <= m; i++){if (v[i] <= limit){add(a[i], b[i], inf);add(b[i], a[i], inf);}}for (int i = 1; i <= n; i++){if (s[i]){add(0, i, s[i]);}if (g[i]){add(i, 201, g[i]);}}int ans = dinic();if (ans == sum)return 1;return 0;
}int main()
{while (scanf("%d", &n) != EOF){if (!n)break;sum = 0;for (int i = 1; i <= n; i++)scanf("%d", s + i), sum += s[i];for (int i = 1; i <= n; i++)scanf("%d", g + i);scanf("%d", &m);int l = 0, r = 1e4;for (int i = 1; i <= m; i++)scanf("%d%d%d", a + i, b + i, v + i);if (!judge(1e4)){printf("No Solution\n");continue;}if (judge(0)){printf("%d\n", 0);continue;}while (r - l > 1){int mid = l + r >> 1;if (judge(mid))r = mid;elsel = mid;}printf("%d\n", r);}
}
C.Function Run Fun
题面
思路
记忆化搜索
示例代码
#include<bits/stdc++.h>using namespace std;#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;ll memo[100][100][100];ll w(int a, int b, int c) {if(a <= 0 || b <= 0 || c <= 0) return 1;if(memo[a][b][c] != -1) return memo[a][b][c];if(a > 20 || b > 20 || c > 20)a = b = c = 20, memo[a][b][c] = w(20, 20, 20);else if(a < b && b < c) memo[a][b][c] = w(a, b, c - 1) + w(a, b - 1, c - 1) - w(a, b - 1, c);else memo[a][b][c] = w(a - 1, b, c) + w(a - 1, b - 1, c) + w(a - 1, b, c - 1) - w(a - 1, b - 1, c - 1);return memo[a][b][c];
}signed main() {//ios::sync_with_stdio(false); cin.tie(nullptr);int a, b, c;memset(memo, -1, sizeof(memo));while(cin >> a >> b >> c) {if(a == -1 && b == -1 && c == -1) break;printf("w(%d, %d, %d) = %lld\n", a, b, c, w(a, b, c));}return 0;
}
D.Equation Solver
题面
示例代码
#include <iostream>
#include <string>
#include <cmath>
#include <iomanip>
using namespace std;const double EPS = 1e-8;pair<double, double> parse_expression(const string& s, int& pos);
pair<double, double> parse_term(const string& s, int& pos);
pair<double, double> parse_factor(const string& s, int& pos);int parse_number(const string& s, int& pos) {int num = 0;while (pos < s.size() && isdigit(s[pos])) {num = num * 10 + (s[pos] - '0');pos++;}return num;
}pair<double, double> parse_factor(const string& s, int& pos) {if (s[pos] == 'x') {pos++;return {1.0, 0.0};} else if (isdigit(s[pos])) {int num = parse_number(s, pos);return {0.0, (double)num};} else if (s[pos] == '(') {pos++;auto res = parse_expression(s, pos);if (pos < s.size() && s[pos] == ')') {pos++;}return res;} else {return {0.0, 0.0};}
}pair<double, double> parse_term(const string& s, int& pos) {auto res = parse_factor(s, pos);double a = res.first, b = res.second;while (pos < s.size() && s[pos] == '*') {pos++;auto factor = parse_factor(s, pos);double new_a = a * factor.second + b * factor.first;double new_b = b * factor.second;a = new_a;b = new_b;}return {a, b};
}pair<double, double> parse_expression(const string& s, int& pos) {pair<double, double> res = {0.0, 0.0};bool positive = true;if (pos < s.size() && (s[pos] == '+' || s[pos] == '-')) {positive = (s[pos] == '+');pos++;}auto term = parse_term(s, pos);if (!positive) {term.first *= -1;term.second *= -1;}res.first = term.first;res.second = term.second;while (pos < s.size() && (s[pos] == '+' || s[pos] == '-')) {char op = s[pos];pos++;auto term = parse_term(s, pos);if (op == '+') {res.first += term.first;res.second += term.second;} else {res.first -= term.first;res.second -= term.second;}}return res;
}int main() {string line;int case_num = 1;while (getline(cin, line)) {int eq_pos = line.find('=');string left_str = line.substr(0, eq_pos);string right_str = line.substr(eq_pos + 1);int pos = 0;auto left = parse_expression(left_str, pos);pos = 0;auto right = parse_expression(right_str, pos);double total_a = left.first - right.first;double total_b = right.second - left.second;cout << "Equation #" << case_num << endl;if (fabs(total_a) < EPS) {if (fabs(total_b) < EPS) {cout << "Infinitely many solutions." << endl;} else {cout << "No solution." << endl;}} else {double x = total_b / total_a;if(abs(x) <1e-9) x = abs(x);cout << "x = " << fixed << setprecision(6) << x << endl;}cout << endl;case_num++;}return 0;
}
E.Triangular Sums
题面
思路
加权求和，直接模拟即可
示例代码
#include<bits/stdc++.h>using namespace std;#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;signed main() {ios::sync_with_stdio(false); cin.tie(nullptr);int t;cin >> t;fer(i, 1, t + 1) {int n;cin >> n;ll ans = 0;fer(k, 1, n + 1) ans += k * (1 + k + 1) * (k + 1) / 2;cout << i << ' ' << n << ' ' << ans << '\n';}return 0;
}
F.W‘s Cipher
题面
思路
将字符串字符分成三组，将每组进行循环移位，注意处理空组串的情况
示例代码
#include<bits/stdc++.h>using namespace std;#define ll long long
//#define int ll
#define pii pair<int, int>
#define all(x) x.begin(),x.end()
#define fer(i, m, n) for(int i = m; i < n; ++i)
#define ferd(i, m, n) for(int i = m; i >= n; --i)
#define dbg(x) cout << #x << ' ' << char(61) << ' ' << x << '\n'const int MOD = 1e9 + 7;
const int N = 2e5 + 2;
const int inf = 1e9;void fun(string &s, int k) {k %= s.size();if(k < 0) k += s.size();s = s.substr(s.size() - k) + s.substr(0, s.size() - k);
}signed main() {ios::sync_with_stdio(false); cin.tie(nullptr);int k1, k2, k3;while(cin >> k1 >> k2 >> k3, k1 || k2 || k3) {string s;cin >> s;vector<int> table(s.size());string s1, s2, s3;fer(i, 0, s.size()) {if(s[i] >= 'a' && s[i] <= 'i') table[i] = 1, s1 += s[i];else if(s[i] >= 'j' && s[i] <= 'r') table[i] = 2, s2 += s[i];else table[i] = 3, s3 += s[i];}if(s1.size()) fun(s1, k1);if(s2.size()) fun(s2, k2);if(s3.size()) fun(s3, k3);string ans;int ind1 = 0, ind2 = 0, ind3 = 0;fer(i, 0, s.size()) {if(table[i] == 1) ans += s1[ind1++];else if(table[i] == 2) ans += s2[ind2++];else ans += s3[ind3++];}cout << ans << '\n';}return 0;
}
