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 10
int 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 10
int x[MAX_K],y[MAX_K];
#define MAX_X 101
#define MAX_Y 101
bool 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 10
int 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 10
int 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);
}
}
}
}
}