首页 > ACM题库 > HDU-杭电 > hdu 2389 Rain on your Parade-分治-[解题报告]C++
2014
01-05

hdu 2389 Rain on your Parade-分治-[解题报告]C++

Rain on your Parade

问题描述 :

You’re giving a party in the garden of your villa by the sea. The party is a huge success, and everyone is here. It’s a warm, sunny evening, and a soothing wind sends fresh, salty air from the sea. The evening is progressing just as you had imagined. It could be the perfect end of a beautiful day.
But nothing ever is perfect. One of your guests works in weather forecasting. He suddenly yells, “I know that breeze! It means its going to rain heavily in just a few minutes!” Your guests all wear their best dresses and really would not like to get wet, hence they stand terrified when hearing the bad news.
You have prepared a few umbrellas which can protect a few of your guests. The umbrellas are small, and since your guests are all slightly snobbish, no guest will share an umbrella with other guests. The umbrellas are spread across your (gigantic) garden, just like your guests. To complicate matters even more, some of your guests can’t run as fast as the others.
Can you help your guests so that as many as possible find an umbrella before it starts to pour?

Given the positions and speeds of all your guests, the positions of the umbrellas, and the time until it starts to rain, find out how many of your guests can at most reach an umbrella. Two guests do not want to share an umbrella, however.

输入:

The input starts with a line containing a single integer, the number of test cases.
Each test case starts with a line containing the time t in minutes until it will start to rain (1 <=t <= 5). The next line contains the number of guests m (1 <= m <= 3000), followed by m lines containing x- and y-coordinates as well as the speed si in units per minute (1 <= si <= 3000) of the guest as integers, separated by spaces. After the guests, a single line contains n (1 <= n <= 3000), the number of umbrellas, followed by n lines containing the integer coordinates of each umbrella, separated by a space.
The absolute value of all coordinates is less than 10000.

输出:

The input starts with a line containing a single integer, the number of test cases.
Each test case starts with a line containing the time t in minutes until it will start to rain (1 <=t <= 5). The next line contains the number of guests m (1 <= m <= 3000), followed by m lines containing x- and y-coordinates as well as the speed si in units per minute (1 <= si <= 3000) of the guest as integers, separated by spaces. After the guests, a single line contains n (1 <= n <= 3000), the number of umbrellas, followed by n lines containing the integer coordinates of each umbrella, separated by a space.
The absolute value of all coordinates is less than 10000.

样例输入:

2
1
2
1 0 3
3 0 3
2
4 0
6 0
1
2
1 1 2
3 3 2
2
2 2
4 4

样例输出:

Scenario #1:
2

Scenario #2:
2

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <string.h>
#include <vector>
#include <queue>
using namespace std;
/* *******************************
 * 二分图匹配(Hopcroft-Carp算法)
 * 复杂度O(sqrt(n)*E)
 * 邻接表存图,vector实现
 * vector先初始化,然后假如边
 * uN 为左端的顶点数,使用前赋值(点编号0开始)
 */
const int MAXN = 3030;
const int INF = 0x3f3f3f3f;
vector<int>G[MAXN];
int uN;

int Mx[MAXN],My[MAXN];
int dx[MAXN],dy[MAXN];
int dis;
bool used[MAXN];
bool SearchP()
{
    queue<int>Q;
    dis = INF;
    memset(dx,-1,sizeof(dx));
    memset(dy,-1,sizeof(dy));
    for(int i = 0 ; i < uN; i++)
        if(Mx[i] == -1)
        {
            Q.push(i);
            dx[i] = 0;
        }
    while(!Q.empty())
    {
        int u = Q.front();
        Q.pop();
        if(dx[u] > dis)break;
        int sz = G[u].size();
        for(int i = 0;i < sz;i++)
        {
            int v = G[u][i];
            if(dy[v] == -1)
            {
                dy[v] = dx[u] + 1;
                if(My[v] == -1)dis = dy[v];
                else
                {
                    dx[My[v]] = dy[v] + 1;
                    Q.push(My[v]);
                }
            }
        }
    }
    return dis != INF;
}
bool DFS(int u)
{
    int sz = G[u].size();
    for(int i = 0;i < sz;i++)
    {
        int v = G[u][i];
        if(!used[v] && dy[v] == dx[u] + 1)
        {
            used[v] = true;
            if(My[v] != -1 && dy[v] == dis)continue;
            if(My[v] == -1 || DFS(My[v]))
            {
                My[v] = u;
                Mx[u] = v;
                return true;
            }
        }
    }
    return false;
}
int MaxMatch()
{
    int res = 0;
    memset(Mx,-1,sizeof(Mx));
    memset(My,-1,sizeof(My));
    while(SearchP())
    {
        memset(used,false,sizeof(used));
        for(int i = 0;i < uN;i++)
            if(Mx[i] == -1 && DFS(i))
                res++;
    }
    return res;
}

struct Point
{
    int x,y,s;
    void input1()
    {
        scanf("%d%d%d",&x,&y,&s);
    }
    void input2()
    {
        scanf("%d%d",&x,&y);
    }
};
int dis2(Point a,Point b)
{
    return (a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y);
}
Point p1[MAXN],p2[MAXN];

int main()
{
    int T;
    int t;
    int iCase = 0;
    int n,m;
    scanf("%d",&T);
    while(T--)
    {
        iCase++;
        scanf("%d",&t);
        scanf("%d",&n);
        for(int i = 0;i < n;i++)
            p1[i].input1();
        scanf("%d",&m);
        for(int i = 0;i < m;i++)
            p2[i].input2();
        for(int i = 0;i < n;i++)
            G[i].clear();
        uN = n;
        for(int i = 0;i < n;i++)
            for(int j = 0;j < m;j++)
                if(dis2(p1[i],p2[j]) <= p1[i].s*p1[i].s*t*t)
                    G[i].push_back(j);
        printf("Scenario #%d:\n",iCase);
        printf("%d\n\n",MaxMatch());
    }
    return 0;
}

解题转自:http://www.cnblogs.com/kuangbin/archive/2013/07/29/3224091.html


  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。

  2. #include <cstdio>
    #include <algorithm>

    struct LWPair{
    int l,w;
    };

    int main() {
    //freopen("input.txt","r",stdin);
    const int MAXSIZE=5000, MAXVAL=10000;
    LWPair sticks[MAXSIZE];
    int store[MAXSIZE];
    int ncase, nstick, length,width, tmp, time, i,j;
    if(scanf("%d",&ncase)!=1) return -1;
    while(ncase– && scanf("%d",&nstick)==1) {
    for(i=0;i<nstick;++i) scanf("%d%d",&sticks .l,&sticks .w);
    std::sort(sticks,sticks+nstick,[](const LWPair &lhs, const LWPair &rhs) { return lhs.l>rhs.l || lhs.l==rhs.l && lhs.w>rhs.w; });
    for(time=-1,i=0;i<nstick;++i) {
    tmp=sticks .w;
    for(j=time;j>=0 && store >=tmp;–j) ; // search from right to left
    if(j==time) { store[++time]=tmp; }
    else { store[j+1]=tmp; }
    }
    printf("%dn",time+1);
    }
    return 0;
    }

  3. 第一句可以忽略不计了吧。从第二句开始分析,说明这个花色下的所有牌都会在其它里面出现,那么还剩下♠️和♦️。第三句,可以排除2和7,因为在两种花色里有。现在是第四句,因为♠️还剩下多个,只有是♦️B才能知道答案。