POJ-3021 e-Coins(广搜)

e-Coins

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 699 Accepted: 348

Description

At the Department for Bills and Coins, an extension of today's monetary system has newly been proposed, in order to make it fit the new economy better. A number of new so called e-coins will be produced, which, in addition to having a value in the normal sense of today, also have an InfoTechnological value. The goal of this reform is, of course, to make justice to the economy of numerous dotcom companies which, despite the fact that they are low on money surely have a lot of IT inside. All money of the old kind will keep its conventional value and get zero InfoTechnological value. 

To successfully make value comparisions in the new system, something called the e-modulus is introduced. This is calculated as SQRT(X*X+Y*Y), where X and Y hold the sums of the conventional and InfoTechnological values respectively. For instance, money with a conventional value of $3 altogether and an InfoTechnological value of $4 will get an e-modulus of $5. Bear in mind that you have to calculate the sums of the conventional and InfoTechnological values separately before you calculate the e-modulus of the money. 

To simplify the move to e-currency, you are assigned to write a program that, given the e-modulus that shall be reached and a list of the different types of e-coins that are available, calculates the smallest amount of e-coins that are needed to exactly match the e-modulus. There is no limit on how many e-coins of each type that may be used to match the given e-modulus.

Input

A line with the number of problems n (0 < n<=100), followed by n times:

• A line with the integers m (0 < m<=40) and S (0 < S<=300), where m indicates the number of different e-coin types that exist in the problem, and S states the value of the e-modulus that shall be matched exactly.
• m lines, each consisting of one pair of non-negative integers describing the value of an e-coin. The first number in the pair states the conventional value, and the second number holds the InfoTechnological value of the coin.

When more than one number are present on a line, they will be separated by a space. Between each problem, there will be one blank line.

Output

The output consists of n lines. Each line contains either a single integer holding the number of coins necessary to reach the specified e-modulus S or, if S cannot be reached, the string "not possible".

Sample Input

3 2 5 0 2 2 0 3 20 0 2 2 0 2 1 3 5 3 0 0 4 5 5 

Sample Output

not possible 10 2

Hint

 
The illustration examplifies adding 8 coins of conventional value 2 and InfoTechnological value 1, and 2 coins with pure InfoTechnological value 2. The e-modulus is of course 20 as SQRT((8*2+2*0)^2+(8*1+2*2)^2)=SQRT(16^2+12^2)=20 

题意：从（0,0）点到距离他为n的点最少需要多少步；

           题目中输入m组数据，表示一步可以走的m种情况。

思路：与一边的广搜题目差不多，只要读懂题意就好，只是广搜向外拓展时是枚举输入的m种情况

           图没有界限，但是要保证拓展的点到（0,0）点的距离小于等于n;

代码：

#include<stdio.h>#include<math.h>#include<string.h>#include<queue>using namespace std;struct node{    int x;    int y;    int step;};int u[200],v[200],m,n;int book[400][400];void bfs(){    node now,tmp;    queue<node> q;    now.x=0;    now.y=0;    now.step=0;    book[0][0]=1;    q.push(now);    while(!q.empty())    {        now=q.front();        q.pop();        if(sqrt(now.x*now.x+now.y*now.y)==n)        {            printf("%d\n",now.step);            return ;        }        for(int i=0;i<m;i++)//枚举m种情况        {            int tx=now.x+u[i];            int ty=now.y+v[i];            if(tx*tx+ty*ty<=n*n&&book[tx][ty]==0)//保证（tx,ty）到（0,0）的距离<=n            {                book[tx][ty] = 1;                tmp.x = tx;                tmp.y = ty;                tmp.step = now.step + 1;//步数                q.push(tmp);            }        }    }    printf("not possible\n");}int main(){    int t;    scanf("%d",&t);    while(t--)    {        scanf("%d%d",&m,&n);//m种走法，n是距离（0,0）的距离        for(int i=0 ; i<m ;i++)            scanf("%d%d",&u[i],&v[i]);        memset(book,0,sizeof(book));//标记是否走过        bfs();    }    return 0;}