南邮 OJ 1966 MOVE

MOVE

时间限制(普通/Java) : 2000 MS/ 10000 MS          运行内存限制 : 81920 KByte
总提交 : 90            测试通过 : 16 

比赛描述

On a infinite plane,you are on the (0,0) at the beginning.
You have to move to (x,y).But you can only move by the give K ways.
I want you to give me the least steps you need to move to the destination.

For example
 If you have to go to (3,1) and you have 3 ways to move: (-1,1) (1,0)(3,1).you only need one (3,1) move to get to the destination.

There are more than one case in a single input.
I promise that you can get to the destination.

输入

In the first there is a T,which means T cases;
In the first line there are 3 integers x y K   -100<x,y<100  k<=10
In the following K line ,each line has 2 integers xi,yi   -100<xi,yi<100;

输出

One integer to show the minimum steps.

样例输入

1
1 1 1
1 1

样例输出

1

题目来源

2E

/* Time Limit Exceed at Test 1#include<iostream>#include<queue>using namespace std;#define MAX_K 10int x[MAX_K],y[MAX_K];struct point{int x,y,step;};int main(){int t,target_x,target_y,k,i;point p,p1;queue<point> qp;scanf("%d%",&t);while(t--){scanf("%d%d%d",&k,&target_x,&target_y);for(i=0;i<k;i++){scanf("%d%d",x+i,y+i);}p.x = p.y = p.step = 0;while(!qp.empty()){qp.pop();}qp.push(p);while(!qp.empty()){p = qp.front();qp.pop();if(p.x==target_x && p.y==target_y){printf("%d\n",p.step);break;}p1.step = p.step+1;for(i=0;i<k;i++){p1.x = p.x+x[i];p1.y = p.y+y[i];}qp.push(p1);}}}*//* Wrong Answer at Test 1#include<iostream>#include<queue>using namespace std;#define MAX_K 10int x[MAX_K],y[MAX_K];#define MAX_X 101#define MAX_Y 101bool vst[MAX_X][MAX_X];struct point{int x,y,step;};int main(){int t,target_x,target_y,k,i;point p,p1;queue<point> qp;scanf("%d%",&t);while(t--){memset(vst,0,sizeof(vst));scanf("%d%d%d",&k,&target_x,&target_y);for(i=0;i<k;i++){scanf("%d%d",x+i,y+i);}p.x = p.y = p.step = 0;while(!qp.empty()){qp.pop();}qp.push(p);vst[p.x][p.y] = 1;while(!qp.empty()){p = qp.front();qp.pop();if(p.x==target_x && p.y==target_y){printf("%d\n",p.step);break;}p1.step = p.step+1;for(i=0;i<k;i++){p1.x = p.x+x[i];p1.y = p.y+y[i];if(0<=p1.x && p1.x<MAX_X && 0<=p1.y && p1.y<MAX_Y && !vst[p1.x][p1.y]){vst[p1.x][p1.y] = 1;qp.push(p1);}}}}}*//* Wrong Answer at Test 1#include<iostream>#include<queue>#include<set>using namespace std;#define MAX_K 10int x[MAX_K],y[MAX_K];struct point{int x,y,step;};bool operator<(const point p1, const point p2){if(p1.x == p2.x){return p1.y < p2.y;}return p1.x < p2.x;}int main(){//freopen("test.txt","r",stdin);int t,target_x,target_y,k,i;point p,p1;queue<point> qp;set<point> sp;scanf("%d%",&t);while(t--){scanf("%d%d%d",&k,&target_x,&target_y);for(i=0;i<k;i++){scanf("%d%d",x+i,y+i);}while(!qp.empty()){qp.pop();}sp.clear();p.x = p.y = p.step = 0;qp.push(p);sp.insert(p);while(!qp.empty()){p = qp.front();qp.pop();if(p.x==target_x && p.y==target_y){printf("%d\n",p.step);break;}p1.step = p.step+1;for(i=0;i<k;i++){p1.x = p.x+x[i];p1.y = p.y+y[i];if(!sp.count(p1)){sp.insert(p1);qp.push(p1);}}}}}*///1906MS#include<iostream>#include<queue>#include<set>using namespace std;#define MAX_K 10int x[MAX_K],y[MAX_K];struct point{int x,y,step;};bool operator<(const point p1, const point p2){if(p1.x == p2.x){return p1.y < p2.y;}return p1.x < p2.x;}int main(){//freopen("test.txt","r",stdin);int t,target_x,target_y,k,i;point p,p1;queue<point> qp;set<point> sp;scanf("%d",&t);while(t--){scanf("%d%d%d",&target_x,&target_y,&k);for(i=0;i<k;i++){scanf("%d%d",x+i,y+i);}while(!qp.empty()){qp.pop();}sp.clear();p.x = p.y = p.step = 0;qp.push(p);sp.insert(p);while(!qp.empty()){p = qp.front();qp.pop();if(p.x==target_x && p.y==target_y){printf("%d\n",p.step);break;}p1.step = p.step+1;for(i=0;i<k;i++){p1.x = p.x+x[i];p1.y = p.y+y[i];if(!sp.count(p1)){sp.insert(p1);qp.push(p1);}}}}}