Regionals 2012 :: HangZhou

题目传送门
排行榜

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;
}

　　

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;
}

　　

水 I The Power of Xiangqi

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;
}

　　

时间： 2024-10-04 01:08:23

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