Typhoon

Time Limit:1000MS  Memory Limit:65536K
Total Submit:43 Accepted:8

Description

Twilight's hometown, Fuzhou, Fujian's provincial capital, is attacked by a number of typhoons every summer.

Every time a typhoon comes, meteorological station will be issued warnings and forecast the typhoon moving line.

If a typhoon affected Fuzhou, twilight would have to stay at home, which made him very depressed.

Therefore, twilight uses the data released by the weather stations to forecast wether a typhoon would affect his home every time.

For simplicity, the map is considered as a two-dimensional plane,at the beginning the typhoon is located in (0,0) point, assuming that the typhoon's path is a straight line and keeps the same radius when it moves , if his home is just on the border of the typhoon he will not be affected.

Input

Each group of data includes five integers, x0, y0, x1, y1, R.
(x0, y0) is twilight's home's coordinates.
(x1, y1) is the direction vector of the typhoon.
R is the radius of typhoon.
-1000 <= X0, y0, x1, y1 <= 1000
x1^2+y1^2 ! = 0 ,0 <R <=500
Five 0 means input is end, you should not deal with this line of data.

Output

If the typhoon will affect the twilight's home, output Yes, otherwise No.
Print a blank line after each test case.

Sample Input

1 0 0 1 1

0 0 0 0 0

Sample Output

No

 该题台风走的是一条射线，假设过原点由方向向量确定的直线为f(x,y),和f(x,y)垂直的直线为f’(x,y),我们分两种情况讨论:

1.           当点位于台风方向向量的相反方向,即在f’(x,y)的另一侧，如果点在以原点为圆心，r为半径的区域内(不包括边界),则受到影响，否则不受到影响.

因为f’(x,y)=x1*x+y1*y ,且当f’(x1,y1)* f’(x0,y0)<0时，（x1,y1）和（x0,y0）位于f’(x,y)的两侧。又因为f’(x1,y1)>0 ,所以本题之判断f’(x0,y0)<0即可

2.           当点位于台风方向向量的同一方向时，只要判断点到直线的距离是否小于r，小于的话就影响，否则不影响。又因为开方的速率要远远低于乘方，所以本题乘方后判断。

Source

/*

 Problem:Typhoon

 Author:Chad

 Algrithem:(x1*y0+y1*x0)^2 >= r^2(x1^2+ y1^2)

 Date:2009-11-25

*/

#include<iostream>

using namespace std;

int x0, y0, x1, y1,r;

int main()

{

    int n=0;

    while(cin>>x0>>y0>>x1>>y1>>r && !(!x0 && !y0 && !x1 && !y1 && !r))

    {

         if((n++)>0) cout<<endl;

         //

         if(x1*x0+y1*y0 < 0)

         {

             if(x1*x1+y1*y1>=r*r) cout<<"No\n";

             else cout<<"Yes\n";

         }

         else if((x1*y0-y1*x0)*(x1*y0-y1*x0) >= r*r*(x1*x1+ y1*y1))

             cout<<"No\n";

         else

             cout<<"Yes\n";

    }

    return 0;

}

这道题目是一道动态规划（DP）题。

状态转移方程为：

（dp[i][j]表示在第i分钟结束时有体力j）

dp[i][j]=dp[i-1][j-1]+d[i] (j>0)

dp[i][0]=max{dp[i-j][j],dp[i-1][0]} (i-j>=j,j<=m)

为什么这样写呢？

后面一个方程比较好理解，j=0肯定是从前面已知休息得来的。

前面一个方程为什么不可能是dp[i][j]=dp[i-1][j+1]呢,因为这种情况已经包含在下面一个方程里了（注意，要休息必须已知休息到j=0）

ｃｏｄｅ

#include <stdio.h>
int dp[10001][501]; 
int n,m,i,j;
int max(int i,int m)
{
        int ma=dp[i-1][1];
        for (int j=2;j<=m && i-j>=j;j++)
            if (dp[i-j][j]>ma) 
                ma=dp[i-j][j];
        if (dp[i-1][0]>ma)
            ma=dp[i-1][0];
        return ma;
}        
int main()
{
    scanf("%d%d",&n,&m);
    int d[10001];
    for (i=1;i<=n;i++)
        scanf("%d",&d[i]);
    for (i=0;i<501;i++)
        dp[i][0]=0;
    for (i=1;i<=n;i++)
    {
        for (j=1;j<=m && j<=i;j++)
            dp[i][j]=dp[i-1][j-1]+d[i];
        dp[i][0]=max(i,m);
    }
    printf("%d\n",dp[n][0]);
    return 0;
}