比赛征程
| # | A | B | C | D | E | F |
|---|---|---|---|---|---|---|
| contest18 | Y | Y | Y | A | Y | A |
A - Plate Game
Problem Description
往N*M的矩阵放置 半径为R的圆，再也不能放则输。
Time Limit: 2s
Solution
显然当且仅当初次就无法放下圆时先手败, 若圆能放在桌子上,那先手就一定能放在桌子的正中间, 不论对手怎么放,先手总能通过与之中心对称放置得到可行解.
Code
#include <cstdio> #include <iostream> using namespace std; int main() { int a, b, r; scanf("%d%d%d", &a, &b, &r); if (a > b) swap(a, b); if (r * 2 > a) puts("Second"); else puts("First"); return 0; }
B - Limit
Problem Description
题意：第一行给出两个数字m,n，每个数字代表两个表达式的最高次项，接下来两行数据分别有表示从x^n-x^0的系数,求当x趋向于正无穷时，这两个表达式比值的极限.
Time Limit: 2s
Solution
很简单的极限判断,若次数相同, 在x 趋向无限大 大时只有最高位有效.
Code
#include <cstdio> #include <iostream> using namespace std; int x[3000], y[3000]; inline int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); } inline int mabs(int x) { if (x < 0) return -x; return x; } int main() { int a, b; scanf("%d%d", &a, &b); a ++; b ++; for (int i = 1; i <= a; i ++) scanf("%d", &x[i]); for (int i = 1; i <= b; i ++) scanf("%d", &y[i]); if (a > b) { if (x[1] * y[1] > 0) puts("Infinity"); else puts("-Infinity"); } else if (b > a) puts("0/1"); else { int tmp = gcd(mabs(x[1]), mabs(y[1])); x[1] /= tmp; y[1] /= tmp; if (x[1] * y[1] > 0) printf("%d/%d\n", mabs(x[1]), mabs(y[1])); else printf("-%d/%d\n", mabs(x[1]), mabs(y[1])); } return 0; }
C - Lexicographically Maximum Subsequence
Problem Description
给出一个字符串，输出子串中字典序最大的一个
Time Limit: 2s
Solution
先排序, 不断的取掉字典序大的字符.
Code
#include <cstdio> #include <iostream> #include <algorithm> #include <cstring> using namespace std; struct node { char c; int id; }q[100005]; char s[100005]; inline bool cmp(node a, node b) { if (a.c != b.c) return a.c > b.c; return a.id < b.id; } int main() { scanf("%s", s); int l = strlen(s); for (int i = 0; i < l; i ++) q[i + 1].c = s[i], q[i + 1].id = i + 1; sort(q + 1, q + l + 1, cmp); int p = 1, mx = 0; while (p <= l) { while (p <= l && q[p].id < mx) p ++; if (p > l) break; mx = q[p].id; printf("%c", q[p].c); p ++; } printf("\n"); return 0; }
D - Infinite Maze
Problem Description
给出无限大的循环迷宫, 问从出发点开始是否能走到无限远.
Time Limit: 2s
Solution
首先此题必然是广搜, 然后重点就是判断的条件了.我们在单位迷宫上走,如果越界就走到另一个单位迷宫上,ppx[x%n][y%m]=x 与 ppy[x%n][y%m]=y记忆化, 若能在不同的单位迷宫上走到同一个相对位置,那么必然可以走到无限远.
Code
#include <cstdio> #include <iostream> #include <cstring> #include <queue> using namespace std; const int N = 1510; struct node { int x, y, px, py; }; char G[N][N]; int ppx[N][N], ppy[N][N]; int n, m, x, y, nx, ny, px, py; int dx[4] = {0, 0, 1, -1}; int dy[4] = {1, -1, 0, 0}; queue<node> Q; int main() { scanf("%d%d", &n, &m); for (int i = 0; i < n; i ++) { scanf("%s", G[i]); for (int j = 0; j < m; j ++) if (G[i][j] == 'S') { x = i; y = j; G[i][j] = '.'; } } memset(ppx, 127, sizeof(ppx)); memset(ppy, 127, sizeof(ppy)); ppx[x][y] = ppy[x][y] = 0; Q.push((node){x, y, 0, 0}); node cur; while (!Q.empty()) { cur = Q.front(); Q.pop(); for (int i = 0; i < 4; i ++) { nx = cur.x + dx[i]; ny = cur.y + dy[i]; px = cur.px; py = cur.py; if (nx < 0) { nx = n - 1; px --; } else if (nx >= n) { nx = 0; px ++; } if (ny < 0) { ny = m - 1; py --; } else if (ny >= m) { ny = 0; py ++; } if (G[nx][ny] == '.') { if (ppx[nx][ny] == ppx[1500][1500]) { ppx[nx][ny] = px; ppy[nx][ny] = py; Q.push((node){nx, ny, px, py}); } else if (ppx[nx][ny] != px || ppy[nx][ny] != py) { puts("Yes"); return 0; } } } } puts("No"); return 0; }
E - Paint Tree
Problem Description
给出一棵树和平面直角坐标系上的点, 求一个树上节点到平面上点的映射, 使得映射后平面上的这棵树上的边不相交,平面上没有三点共线.
Time Limit: 2s
Solution
贪心题,贪心的直接目的是使得对于一个节点, 它的子树之间互不相交.那么我们可以递归处理,先初始化得到每个节点的子树大小, 对于一棵子树和一些选定的点的映射, 可以取点中最下方的点与子树的根映射, 然后将剩下的点按照极角排序, 然后根据该点每个子节点的子树的大小分配每个子树的点的区间, 这样根据极角排序可以使每个子树分开,这样就能保证得到可行解.
Code
#include <cstdio> #include <iostream> #include <cmath> #include <algorithm> using namespace std; typedef long long LL; struct edge{ LL nxt, to; }e[3005]; int cnt = 0, fir[1505], size[1505]; struct point{ LL x; LL y; int id; }P[1505], bb; int ans[1505]; inline bool cmp(point a, point b) { return ((a.x - bb.x) * (b.y - bb.y) - (a.y - bb.y) * (b.x - bb.x)) > 0; } inline void add(int x, int y) { e[++ cnt].to = y; e[cnt].nxt = fir[x]; fir[x] = cnt; } inline void getsize(int x, int pa) { size[x] = 1; for (LL i = fir[x]; i; i = e[i].nxt) if (e[i].to != pa) { getsize(e[i].to, x); size[x] += size[e[i].to]; } } inline void solve(int l, int r, int rt, int pa) { if (l > r) return; if (l == r) { ans[P[l].id] = rt; return; } LL mx = 2e9, my = 2e9; int rec = 0; for (LL i = l; i <= r; i ++) if (P[i].y < my || (P[i].y == my && P[i].x < mx)) { mx = P[i].x; my = P[i].y; rec = i; } swap(P[rec], P[l]); ans[P[l].id] = rt; bb = P[l]; sort(P + l + 1, P + r + 1, cmp); int pp = l + 1, np; for (int i = fir[rt]; i; i = e[i].nxt) if (e[i].to != pa) { np = pp + size[e[i].to]; solve(pp, np - 1, e[i].to, rt); pp = np; } } int main() { int n, x, y; scanf("%d", &n); for (int i = 1; i < n; i ++) { scanf("%d%d", &x, &y); add(x, y); add(y, x); } getsize(1, 0); for (int i = 1; i <= n; i ++) scanf("%I64d%I64d", &P[i].x, &P[i].y), P[i].id = i; solve(1, n, 1, 0); for (int i = 1; i <= n; i ++) printf("%d%s", ans[i], i == n ? "\n" : " "); }
F - Opening Portals
Problem Description
给一张图,从一号点出发, 图上有一些点初始状态为0,若某两个点状态都为1则可以瞬间转移,到达一个点时若它有状态且为0则瞬间改为1,求把所有状态为0的点变成1所经过的最小路程.
Time Limit: 2s
Solution
如果1号点没有状态, 那么首先必定是走到一个最近的有状态的点上, 然后利用瞬间转移的性质,题目就变成了求所有有状态点之间的最小生成树,我们把所有的有状态点推入队列中同时跑SPFA并记录每个点由哪个有状态的点(记为fro)转移来,那么我们得到的就是每个点到最近的有状态节点的距离(dis), 对于原树上的连接i, j的权为wei的边就相当于有长度为dis[i]+dis[j]+wei的边连接fro[i]与fro[j], 以此进行kruskal即可.
Code
#include <cstdio> #include <iostream> #include <queue> #include <cstring> #include <algorithm> using namespace std; typedef long long LL; const int N = 100005; struct edge{ int fro, to, nxt; LL c; }e[N * 2]; int cnt = 0, fir[N]; int n, m, x, y, k; int port[N], back[N]; int fro[N]; int fa[N]; LL dis[N]; bool inq[N]; queue<int> Q; inline void add(int x, int y, LL l) { e[++ cnt].to = y; e[cnt].fro = x; e[cnt].nxt = fir[x]; e[cnt].c = l; fir[x] = cnt; } inline bool cmp(edge a, edge b) { return a.c + dis[a.fro] + dis[a.to] < b.c + dis[b.fro] + dis[b.to]; } inline int getf(int x) { if (fa[x] == x) return x; return fa[x] = getf(fa[x]); } int main() { LL l; scanf("%d%d", &n, &m); for (int i = 1; i <= m; i ++) { scanf("%d%d%I64d", &x, &y, &l); add(x, y, l); add(y, x, l); } scanf("%d", &k); for (int i = 1; i <= k; i ++) scanf("%d", &port[i]); memset(dis, 127, sizeof(dis)); for (int i = 1; i <= k; i ++) { Q.push(port[i]); dis[port[i]] = 0; fro[port[i]] = port[i]; } while (!Q.empty()) { x = Q.front(); inq[x] = 0; Q.pop(); for (int i = fir[x]; i; i = e[i].nxt) { if (dis[x] + e[i].c < dis[e[i].to]) { dis[e[i].to] = dis[x] + e[i].c; fro[e[i].to] = fro[x]; if (!inq[e[i].to]) { inq[e[i].to] = 1; Q.push(e[i].to); } } } } LL ans = dis[1]; for (int i = 1; i <= n; i ++) fa[i] = i; int u, v; sort(e + 1, e + cnt + 1, cmp); for (int i = 1; i <= cnt; i ++) { u = getf(fro[e[i].fro]); v = getf(fro[e[i].to]); if (u == v) continue; fa[u] = v; ans += dis[e[i].fro] + dis[e[i].to] + e[i].c; } printf("%I64d\n", ans); return 0; }