Regionals 2012 :: HangZhou
一个人做了12年北大出的题，自己还是太弱了，图论的知识忘光光，最小生成树裸题写不来，Dijkstra TLE不知道用SPFA。
简单几何(点到线段的距离) + 三分 B Stealing a Cake
题意：圆外一个点先到圆再到矩形的最短距离。
分析：由于圆在[0, PI]和[PI, PI*2]上是单峰函数，用三分求极值，应该在[0，PI*2]上也算单峰函数吧。
/************************************************
* Author :Running_Time
* Created Time :2015/11/7 星期六 17:24:32
* File Name :B_2.cpp
************************************************/ #include <cstdio>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <cstring>
#include <cmath>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <list>
#include <map>
#include <set>
#include <bitset>
#include <cstdlib>
#include <ctime>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e5 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
int dcmp(double x) { //三态函数，减少精度问题
if (fabs (x) < EPS) return 0;
else return x < 0 ? -1 : 1;
}
struct Point { //点的定义
double x, y;
Point () {}
Point (double x, double y) : x (x), y (y) {}
Point operator + (const Point &r) const { //向量加法
return Point (x + r.x, y + r.y);
}
Point operator - (const Point &r) const { //向量减法
return Point (x - r.x, y - r.y);
}
Point operator * (double p) const { //向量乘以标量
return Point (x * p, y * p);
}
Point operator / (double p) const { //向量除以标量
return Point (x / p, y / p);
}
bool operator < (const Point &r) const { //点的坐标排序
return x < r.x || (x == r.x && y < r.y);
}
bool operator == (const Point &r) const { //判断同一个点
return dcmp (x - r.x) == 0 && dcmp (y - r.y) == 0;
}
};
typedef Point Vector; //向量的定义
Point read_point(void) { //点的读入
double x, y;
scanf ("%lf%lf", &x, &y);
return Point (x, y);
}
double dot(Vector A, Vector B) { //向量点积
return A.x * B.x + A.y * B.y;
}
double cross(Vector A, Vector B) { //向量叉积
return A.x * B.y - A.y * B.x;
}
Vector rotate(Vector A, double rad) {
return Vector (A.x * cos (rad) - A.y * sin (rad), A.x * sin (rad) + A.y * cos (rad));
}
double length(Vector A) {
return sqrt (dot (A, A));
}
double angle(Vector A, Vector B) { //向量转角，逆时针，点积
return acos (dot (A, B) / length (A) / length (B));
}
double point_to_seg(Point p, Point a, Point b) {
if (a == b) return length (p - a);
Vector V1 = b - a, V2 = p - a, V3 = p - b;
if (dcmp (dot (V1, V2)) < 0) return length (V2);
else if (dcmp (dot (V1, V3)) > 0) return length (V3);
else return fabs (cross (V1, V2)) / length (V1);
} struct Circle {
Point c;
double r;
Circle () {}
Circle (Point c, double r) : c (c), r (r) {}
Point point(double a) {
return Point (c.x + cos (a) * r, c.y + sin (a) * r);
}
}; Point s, a, b, c, d;
Circle C; double cal_dis(double rad) {
Point p = C.point (rad);
double dis1 = length (s - p);
double dis2 = 1e9;
dis2 = min (dis2, point_to_seg (p, a, b));
dis2 = min (dis2, point_to_seg (p, b, c));
dis2 = min (dis2, point_to_seg (p, c, d));
dis2 = min (dis2, point_to_seg (p, d, a));
return dis1 + dis2;
} int main(void) {
double x, y, r, x2, y2;
while (scanf ("%lf%lf", &x, &y) == 2) {
if (dcmp (x) == 0 && dcmp (y) == 0) break;
s = Point (x, y);
scanf ("%lf%lf%lf", &x, &y, &r);
C = Circle (Circle (Point (x, y), r));
scanf ("%lf%lf%lf%lf", &x, &y, &x2, &y2);
a = Point (min (x, x2), max (y, y2));
b = Point (min (x, x2), min (y, y2));
c = Point (max (x, x2), min (y, y2));
d = Point (max (x, x2), max (y, y2));
Vector V[2];
int cnt = point_cir_tan (s, C, V);
double l = angle (Vector (1, 0), V[0]), r = angle (Vector (1, 0), V[1]);
if (l > r) swap (l, r);
while (r - l > EPS) {
double mid = (l + r) / 2;
double lmid = (l + mid) / 2;
double rmid = (r + mid) / 2;
double res1 = cal_dis (lmid),
res2 = cal_dis (rmid);
if (res1 < res2) {
r = rmid;
}
else l = lmid;
}
printf ("%.2f\n", cal_dis (l));
}
return 0;
}
预处理+递推DP C Substrings
题意：给n个数字，q次询问，每次回答所有长度为w的子串不同数字个数的和
分析：先想状态，dp[w]表示长度为w的不同数字总和，有个递推关系，dp[i] = dp[i-1] - last[i-1] + sum[i]。last[i]表示后缀长度为i的不同个数，从i-1到i时能发现少了最后一个长度i-1的子串，当然还要加上新的权值，sum[i]表示两个相同数字距离大于等于i的所有个数，那么先求c[i] 等于i的所有个数，后缀累加就行了。这题充分使用了预处理的技巧求解问题，复杂度 O (n)
/************************************************
* Author :Running_Time
* Created Time :2015/11/9 星期一 10:23:58
* File Name :C.cpp
************************************************/ #include <bits/stdc++.h>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e6 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
ll dp[N];
int a[N], pre[N], sum[N], last[N], c[N]; int main(void) {
int n, q;
while (scanf ("%d", &n) == 1) {
if (!n) break;
for (int i=1; i<=n; ++i) scanf ("%d", &a[i]);
memset (c, 0, sizeof (c));
memset (pre, 0, sizeof (pre));
for (int i=1; i<=n; ++i) {
c[i-pre[a[i]]]++;
pre[a[i]] = i;
}
sum[n] = c[n];
for (int i=n-1; i>=1; --i) {
sum[i] = sum[i+1] + c[i];
}
memset (c, 0, sizeof (c));
last[1] = 1; c[a[n]]++;
for (int i=2; i<=n; ++i) { //length
if (c[a[n-i+1]] == 0) {
last[i] = last[i-1] + 1;
c[a[n-i+1]] = 1;
}
else last[i] = last[i-1];
}
dp[1] = n;
for (int i=2; i<=n; ++i) {
dp[i] = dp[i-1] - last[i-1] + sum[i];
}
scanf ("%d", &q);
while (q--) {
int w; scanf ("%d", &w);
printf ("%I64d\n", dp[w]);
}
} //cout << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n"; return 0;
}
SPFA/BFS H Friend Chains
题意：就是求任意两点的最短路的最大值
分析：Floyd O(n^3)绝对超时，我想了Dijkstra，O (V * (VlogE + V)，后来用了SPFA倒是勉勉强强过了，对Dijkstra无爱了。杭电discuss里有人写了最快的解法，类似于求树的直径，两边BFS。问题是第一次找最远的点可能有多个，要选择度数最小的，否则度数多了，从那个点出发到某些点并不是最远的，也就不是直径。
SPFA
#include <bits/stdc++.h>
using namespace std; const int N = 1e3 + 10;
const int E = 2e4 + 10;
const int INF = 0x3f3f3f3f;
struct Edge {
int v, w, nex;
Edge () {}
Edge (int v, double w, int nex) : v (v), w (w), nex (nex) {}
}edge[E];
int head[N];
bool vis[N];
int d[N];
int n, m, e; void init(void) {
memset (head, -1, sizeof (head));
e = 0;
}
void add_edge(int u, int v, int w) {
edge[e] = Edge (v, w, head[u]);
head[u] = e++;
} void SPFA(int s) {
for (int i=1; i<=n; ++i) {
vis[i] = false; d[i] = INF;
}
vis[s] = true; d[s] = 0;
queue<int> Q; Q.push (s);
while (!Q.empty ()) {
int u = Q.front (); Q.pop ();
vis[u] = false;
for (int i=head[u]; ~i; i=edge[i].nex) {
int v = edge[i].v, w = edge[i].w;
if (d[v] > d[u] + w) {
d[v] = d[u] + w;
if (!vis[v]) {
vis[v] = true; Q.push (v);
}
}
}
}
} map<string, int> id;
char str1[11], str2[11]; int main(void) {
while (scanf ("%d", &n) == 1) {
if (!n) break;
id.clear ();
for (int i=1; i<=n; ++i) {
scanf ("%s", &str1);
id[str1] = i;
}
init ();
scanf ("%d", &m);
for (int i=1; i<=m; ++i) {
scanf ("%s%s", &str1, &str2);
if (id[str1] == id[str2]) continue;
add_edge (id[str1], id[str2], 1);
add_edge (id[str2], id[str1], 1);
} int ans = 0;
for (int i=1; i<=n; ++i) {
SPFA (i);
for (int j=1; j<=n; ++j) {
if (d[j] == INF) {
ans = -1; break;
}
if (ans < d[j]) ans = d[j];
}
if (ans == -1) break;
}
printf ("%d\n", ans);
} return 0;
}
BFS
/************************************************
* Author :Running_Time
* Created Time :2015/11/7 星期六 16:16:25
* File Name :H_2.cpp
************************************************/ #include <bits/stdc++.h>
using namespace std; #define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
typedef long long ll;
const int N = 1e3 + 10;
const int E = 2e4 + 10;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-10;
const double PI = acos (-1.0);
struct Edge {
int v, w, nex;
Edge () {}
Edge (int v, int w, int nex) : v (v), w (w), nex (nex) {}
}edge[E];
int head[N], d[N], deg[N];
bool vis[N];
int n, m, e; void init(void) {
memset (head, -1, sizeof (head));
memset (deg, 0, sizeof (deg));
e = 0;
}
void add_edge(int u, int v, int w) {
edge[e] = Edge (v, w, head[u]);
head[u] = e++;
} int BFS(int s) {
for (int i=1; i<=n; ++i) {
d[i] = (i == s) ? 0 : INF;
vis[i] = (i == s) ? true : false;
}
queue<int> Q; Q.push (s);
int idx = 0, mx = 0;
while (!Q.empty ()) {
int u = Q.front (); Q.pop ();
for (int i=head[u]; ~i; i=edge[i].nex) {
int v = edge[i].v, w = edge[i].w;
if (vis[v]) continue;
if (d[v] > d[u] + w) {
d[v] = d[u] + w;
vis[v] = true; Q.push (v);
}
if (d[v] > mx || (d[v] == mx && deg[v] < deg[idx])) {
mx = d[v]; idx = v;
}
}
}
return idx;
}
map<string, int> id;
char str1[11], str2[11]; int main(void) {
while (scanf ("%d", &n) == 1) {
if (!n) break;
id.clear ();
for (int i=1; i<=n; ++i) {
scanf ("%s", &str1);
id[str1] = i;
}
init ();
scanf ("%d", &m);
for (int i=1; i<=m; ++i) {
scanf ("%s%s", &str1, &str2);
if (id[str1] == id[str2]) continue;
add_edge (id[str1], id[str2], 1);
add_edge (id[str2], id[str1], 1);
deg[id[str1]]++; deg[id[str2]]++;
}
int ans = 0;
int idx = BFS (1);
ans = d[BFS (idx)];
for (int i=1; i<=n; ++i) {
if (d[i] == INF) {
ans = -1; break;
}
}
printf ("%d\n", ans);
} return 0;
}
比赛前走了一盘象棋，竟然就碰到了关于象棋的题目。
状态压缩+枚举 J Scaring the Birds
题意：有几个点可以放稻草人，问最少放几个稻草人才能保护所以稻草
分析：没什么陷阱吧，主要看样例的R的范围有些纠结，放稻草人的点没有稻草的。只要状压枚举要放的数量和位置就可以了。
#include <bits/stdc++.h>
using namespace std; const int N = 55;
const int INF = 0x3f3f3f3f;
struct Point {
int x, y, r, step;
Point () {}
Point (int x, int y, int step) : x (x), y (y), step (step) {}
}p[11];
int n, m;
int a[N][N];
bool vis[N][N];
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1}; void map_init(void) {
for (int i=1; i<=n; ++i) {
for (int j=1; j<=n; ++j) a[i][j] = 0;
}
for (int i=0; i<m; ++i) {
a[p[i].x][p[i].y] = 1;
}
} bool judge(void) {
for (int i=1; i<=n; ++i) {
for (int j=1; j<=n; ++j) {
if (!a[i][j]) return false;
}
}
return true;
} bool check(int x, int y) {
if (x < 1 || x > n || y < 1 || y > n || vis[x][y]) return false;
else return true;
} void BFS(int u) {
int R = p[u].r; vis[p[u].x][p[u].y] = true;
queue<Point> Q; Q.push (Point (p[u].x, p[u].y, 0));
while (!Q.empty ()) {
Point r = Q.front (); Q.pop ();
if (r.step > R) continue;
a[r.x][r.y] = 1;
for (int i=0; i<4; ++i) {
int tx = r.x + dx[i], ty = r.y + dy[i];
if (!check (tx, ty)) continue;
vis[tx][ty] = true;
Q.push (Point (tx, ty, r.step + 1));
}
}
} int main(void) {
while (scanf ("%d", &n) == 1) {
if (!n) break;
scanf ("%d", &m);
for (int i=0; i<m; ++i) {
scanf ("%d%d", &p[i].x, &p[i].y);
}
for (int i=0; i<m; ++i) {
scanf ("%d", &p[i].r);
}
int S = 1 << m;
int ans = INF;
for (int i=0; i<S; ++i) {
int num = __builtin_popcount (i);
map_init ();
for (int j=0; j<m; ++j) {
if (i & (1 << j)) {
memset (vis, false, sizeof (vis));
BFS (j);
}
}
if (!judge ()) continue;
ans = min (ans, num);
}
if (ans == INF) puts ("-1");
else printf ("%d\n", ans);
} return 0;
}
最小生成树 K Outlets
题意：求最小生成树，限制是p和q一定要有边连
分析：作为第二水的题目我竟然不会写，kruskal的方法前先将p和q连上边就行了，或者Prim在p到q的距离变为0(优先出队列)，然后再加回去。
Kruskal
#include <bits/stdc++.h>
using namespace std; const int N = 55;
const int E = N * N;
const double INF = 1e9;
struct Edge {
int u, v;
double w;
Edge () {}
Edge (int u, int v, double w) : u (u), v (v), w (w) {}
bool operator < (const Edge &r) const {
return w < r.w;
}
}edge[E];
struct UF {
int rt[N], rk[N];
void init(void) {
memset (rt, -1, sizeof (rt));
memset (rk, 0, sizeof (rk));
}
int Find(int x) {
return rt[x] == -1 ? x : rt[x] = Find (rt[x]);
}
void Union(int x, int y) {
x = Find (x); y = Find (y);
if (x == y) return ;
if (rk[x] > rk[y]) {
rt[y] = x; rk[x] += rk[y];
}
else {
rt[x] = y; rk[y] += rk[x];
}
}
}uf;
int x[N], y[N];
int tot; double sqr(int x, int y) {
return (double) (x * x + y * y);
}
double cal_dis(int i, int j) {
return sqrt (sqr (x[i] - x[j], y[i] - y[j]));
} void add_edge(int u, int v, double w) {
edge[tot++] = Edge (u, v, w);
} int p, q; int main(void) {
int n;
while (scanf ("%d", &n) == 1) {
if (!n) break;
scanf ("%d%d", &p, &q);
if (p > q) swap (p, q);
for (int i=1; i<=n; ++i) {
scanf ("%d%d", &x[i], &y[i]);
}
uf.init (); tot = 0;
double ans = cal_dis (p, q);
uf.rt[p] = q;
for (int i=1; i<=n; ++i) {
for (int j=i+1; j<=n; ++j) {
add_edge (i, j, cal_dis (i, j));
}
}
sort (edge, edge+tot);
for (int i=0; i<tot; ++i) {
int u = edge[i].u, v = edge[i].v;
double w = edge[i].w;
u = uf.Find (u); v = uf.Find (v);
if (u == v) continue;
ans += w;
uf.Union (u, v);
}
printf ("%.2f\n", ans);
} return 0;
}
Prim
#include <bits/stdc++.h>
using namespace std; const int N = 55;
const int E = N * N;
const double INF = 1e9;
struct Edge {
int v, nex;
double w;
Edge () {}
Edge (int v, double w, int nex) : v (v), w (w), nex (nex) {}
bool operator < (const Edge &r) const {
return w > r.w;
}
}edge[E];
int head[N];
bool vis[N];
double d[N];
int x[N], y[N];
int p, q, e; void init(void) {
memset (head, -1, sizeof (head));
e = 0;
} void add_edge(int u, int v, double w) {
edge[e] = Edge (v, w, head[u]);
head[u] = e++;
} double sqr(int x, int y) {
return (double) (x * x + y * y);
}
double cal_dis(int i, int j) {
return sqrt (sqr (x[i] - x[j], y[i] - y[j]));
} double Prim(int s) {
memset (vis, false, sizeof (vis));
for (int i=0; i<55; ++i) d[i] = INF;
priority_queue<Edge> Q;
double ret = 0;
for (int i=head[s]; ~i; i=edge[i].nex) {
int v = edge[i].v; double w = edge[i].w;
if (d[v] > w) {
d[v] = w;
if (v == q) {
ret += d[v];
d[v] = 0;
}
Q.push (Edge (v, d[v], 0));
}
}
vis[s] = true; d[s] = 0;
while (!Q.empty ()) {
int u = Q.top ().v; Q.pop ();
if (vis[u]) continue;
vis[u] = true; ret += d[u];
for (int i=head[u]; ~i; i=edge[i].nex) {
int v = edge[i].v; double w = edge[i].w;
if (v == q) continue;
if (!vis[v] && d[v] > w) {
d[v] = w; Q.push (Edge (v, d[v], 0));
}
}
}
return ret;
} int main(void) {
int n;
while (scanf ("%d", &n) == 1) {
if (!n) break;
scanf ("%d%d", &p, &q);
if (p > q) swap (p, q);
for (int i=1; i<=n; ++i) {
scanf ("%d%d", &x[i], &y[i]);
}
init ();
for (int i=1; i<=n; ++i) {
for (int j=i+1; j<=n; ++j) {
add_edge (i, j, cal_dis (i, j));
add_edge (j, i, cal_dis (i, j));
}
}
printf ("%.2f\n", Prim (p));
} return 0;
}
Regionals 2012 :: HangZhou的更多相关文章
- Regionals 2012 :: Chengdu
题目连接 排行榜 A和I都是签到题 按位BFS K Yet Another Multiple Problem 题意:给一些可以用的数字,求最小的数,它由特定的数字组成且是n的倍数 分析:暴力枚举不可行 ...
- Regionals 2012 :: Asia - Dhaka
水 B Wedding of Sultan 题意:求每个点的度数 分析:可以在,每个字母的的两个端点里求出的的出度,那么除了起点外其他点还有一个入度,再+1 /******************** ...
- 组队练习赛（Regionals 2012, North America - East Central NA）
A.Babs' Box Boutique 给定n个盒子,每个盒子都有长宽高(任意两个盒子长宽高不完全相同),现在选盒子的任意两面,要求x1 <= x2 && y1 <= y ...
- [Regionals 2012 :: Asia - Tokyo ]
链接: https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&category=56 ...
- Regionals 2012, Asia - Jakarta 解题报告
啥都不会做了.. 做题慢死 A.Grandpa's Walk 签到题. 直接DFS就行. 注意先判断这个点可以作为一个路径的起点不. 然后再DFS. 否则处理起来略麻烦 #include <io ...
- Regionals 2012, North America - Greater NY 解题报告
这套题..除了几何的都出了 完全没时间学几何.杯具 A,B,J 水题不解释 C.Pen Counts 这题的话 写几个不等式限制边得范围就行了 然后枚举最小边 D.Maximum Random Wal ...
- 130825组队赛-Regionals 2012, North America - East Central NA
A.Babs' Box Boutique 一道简单的dfs搜索题,需要两两比较,然后搜到底,得到最大值就行了.比赛时队友写的,我只负责debug..赛后自己写的.. #include<iostr ...
- UVALive 6177 The King's Ups and Downs
The King's Ups and Downs Time Limit: 3000ms Memory Limit: 131072KB This problem will be judged on UV ...
- 2012 Asia Hangzhou Regional Contest
Friend Chains http://acm.hdu.edu.cn/showproblem.php?pid=4460 图的最远两点距离,任意选个点bfs,如果有不能到的点直接-1.然后对于所有距离 ...
随机推荐
- 在spring中获取代理对象代理的目标对象工具类
昨天晚上一哥们需要获取代理对象的目标对象,查找了文档发现没有相应的工具类,因此自己写了一个分享给大家.能获取JDK动态代理/CGLIB代理对象代理的目标对象. 问题描述:: 我现在遇到个棘手的问题,要 ...
- java笔记--关于线程死锁
关于线程死锁 什么是死锁: 在编写多线程的时候,必须要注意资源的使用问题,如果两个或多个线程分别拥有不同的资源, 而同时又需要对方释放资源才能继续运行时,就会发生死锁. 简单来说:死锁就是当一个或多个 ...
- java笔记--关于线程通信
关于线程通信 使用多线程编程的一个重要原因就是线程间通信的代价比较小 --如果朋友您想转载本文章请注明转载地址"http://www.cnblogs.com/XHJT/p/3897773.h ...
- Xcode 5.0.2 下载地址
下载地址:http://adcdownload.apple.com/Developer_Tools/xcode_5.0.2/xcode_5.0.2.dmg command_line_tools_os_ ...
- cocos2d调度器（定时执行某函数）
调度器(scheduler) 继承关系 原理介绍 Cocos2d-x调度器为游戏提供定时事件和定时调用服务.所有Node对象都知道如何调度和取消调度事件,使用调度器有几个好处: 每当Node不再可见或 ...
- [BZOJ1998][Hnoi2010]Fsk物品调度
[BZOJ1998][Hnoi2010]Fsk物品调度 试题描述 现在找工作不容易,Lostmonkey费了好大劲才得到fsk公司基层流水线操作员的职位.流水线上有n个位置,从0到n-1依次编号,一开 ...
- nginx学习（一）：基本安装
转载自http://summervast.blog.51cto.com/690507/385511 注意:可能因版本不同,个别指令不起作用,需要注意版本灵活安装,我在安装时也遇到过此问题 开始学习ng ...
- 【Network】一张图看懂 Reactor 与 Proactor 模型的区别
首先来看看Reactor模式,Reactor模式应用于同步I/O的场景.我们以读操作为例来看看Reactor中的具体步骤:读取操作:1. 应用程序注册读就需事件和相关联的事件处理器2. 事件分离器等待 ...
- C++类编程（一）const的使用
设计类时,考虑以下五点 1.构造函数初始化列表 2.函数该不该加const 3.参数传递尽量考虑用引用传递,考虑加不加const 4.返回用不用引用 5.数据尽量放在private,函数尽量放在pub ...
- php配置文件语法
2013年12月11日 09:21:58 ; The syntax of the file is extremely simple. Whitespace and lines ; beginning ...