zoj3761（BFS）

地址：http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5206

Easy billiards

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Time Limit: 2 Seconds       Memory Limit: 65536 KB       Special Judge

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Edward think a game of billiards is too long and boring. So he invented a new game called Easy billiards.

Easy billiards has N balls on a brimless rectangular table in the beginning, and your goal is try to make the number of balls on the table as least as possible by several hit under the following rules:

1: The direction you hit the balls should parallel to the tables border.

2: If ball A crashed into ball B, ball B will moves in the same direction of ball A before the crashing, and ball A will stop in the place of ball B before the crashing.

3: If ball C is moving and there are no balls in front of ball C, it will runs out of the tables border, that means ball C is out of the table.

4: You can choose arbitrary ball on the table to hit, but on a hit, you can't let the ball you choose to hit runs out of the tables border. In another word, a ball could runs out of the table if and only if it was crashed by another ball in a hitting.

Now, Edward wants to know the least number of balls remained on the table after several hits, and how.

Input

There are multiple test cases. For each test cases, in the first line, there is an integer N, which means the number of the balls on the table. There are following N lines, each line contains two integers Xi and Yi, which means the coordinate of ball I. (0<=N<=2000, 0<=Xi, Yi<=10^8)

Output

For each test cases, you should output the least number of balls on the first line. And you should output several lines to show the order of hits following the first line, each line should contains the coordinate of the ball you choose to hit and the direction you hit. (LEFT,RIGHT,UP,DOWN).

Sample Input

4
0 0
2 0
4 0
2 2
9
1 1
2 1
3 1
1 2
2 2
3 2
1 3
2 3
3 3

Sample output

1
(2, 2) DOWN
(4, 0) LEFT
(2, 0) LEFT
1
(1, 3) DOWN
(1, 2) DOWN
(2, 3) DOWN
(2, 2) DOWN
(3, 3) DOWN
(3, 2) DOWN
(3, 1) LEFT
(2, 1) LEFT

题意：一个无边框桌子上有N个球。球可以沿水平或是垂直线移动直到掉下桌面或是撞到别的球，若撞到别的球，那么被其撞到的球将会沿着该方向继续前进。球只可以在有别的球阻碍的方向上移动，问最少桌面可以留下几个球。

思路：用广搜，所以搜一个球，将可以连接到的球全部撞向它，然后将它可以连接到的球能够连接的球再撞向它可以连接到的球，以此类推。A的时候有些某明奇妙，一开始用两个FOR循环现将最少剩下的球的数目输出来，并将可以相连的球都归为一类，然后搜索路径，但不知为什么一直WA。实在无奈，就简化下代码，在搜索时顺便求最小剩下的球的数目，没想到这样就A了。

代码：

AC代码（简化代码）：

#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define LL __int64
#define Max 2222
struct point{
    int x,y,bj;
} pp[2010];
int n;
bool vi[2010];
vector ans;  //用VECTOR记录路径，最后倒着输出
bool kmp(point a,point b){
    return a.bj qi;
    qi.push(here);
    vi[here]=0;
    while(!qi.empty()){
        here=qi.front();
        qi.pop();
        int fap[10]= {0};
        for(int i=1; i<=n; i++){  //这里是判断与其相连的最近的球
            if(pp[i].x==pp[here].x&&pp[i].ypp[here].y){
                if(!fap[2]) fap[2]=i;
                else if(pp[i].y-pp[here].ypp[here].x&&pp[i].y==pp[here].y){
                if(!fap[4]) fap[4]=i;
                else if(pp[i].x-pp[here].x0){
        for(int i=1; i<=n; i++)
            scanf("%d%d",&pp[i].x,&pp[i].y);
        int len=0;
        memset(vi,1,sizeof(vi));
        ans.clear();
        for(int i=1; i<=n; i++)
            if(vi[i]){
                bfs(i);
                len++;
            }
        printf("%d\n",len);
        for(int i=ans.size()-1; i>=0; i--){
            printf("(%d, %d) ",ans[i].x,ans[i].y);
            if(ans[i].bj==1) puts("UP");
            else if(ans[i].bj==2) puts("DOWN");
            else if(ans[i].bj==3) puts("RIGHT");
            else puts("LEFT");
        }
    }
    return 0;
}

WA代码（原先的复杂代码）：

#include
#include
#include
#include
#include
#include
#include
using namespace std;
#define LL __int64
#define Max 2222
struct point
{
    int x,y,bj;
} pp[2010];
int n;
bool vi[2010];
bool kmp(point a,point b)
{
    return a.bj qi;
    vector ans;
    qi.push(here);
    vi[here]=0;
    while(!qi.empty())
    {
        here=qi.front();
        qi.pop();
        int fap[10]= {0};
        for(int i=1; i<=n; i++)
        {
            if(pp[i].bjpp[here].bj) break;
            if(pp[i].x==pp[here].x&&pp[i].ypp[here].y)
            {
                if(!fap[2]) fap[2]=i;
                else if(pp[i].y-pp[here].ypp[here].x&&pp[i].y==pp[here].y)
            {
                if(!fap[4]) fap[4]=i;
                else if(pp[i].x-pp[here].x=0;i--)
    {
        printf("(%d, %d) ",ans[i].x,ans[i].y);
        if(ans[i].bj==1) puts("UP");
        else if(ans[i].bj==2) puts("DOWN");
        else if(ans[i].bj==3) puts("RIGHT");
        else puts("LEFT");
    }
}
int main()
{
    pp[0].bj=0;
    while(scanf("%d",&n)>0)
    {
        int len=1;
        for(int i=1; i<=n; i++)
        {
            scanf("%d%d",&pp[i].x,&pp[i].y);
            pp[i].bj=Max;
        }
        for(int i=1; i<=n; i++)
        {
            if(pp[i].bj==Max) pp[i].bj=++len;
            for(int j=1; j<=n; j++)
            {
                if(pp[i].x==pp[j].x||pp[i].y==pp[j].y)
                {
                    pp[i].bj=min(pp[i].bj,pp[j].bj);
                    pp[j].bj=min(pp[i].bj,pp[j].bj);
                }
            }
        }
        sort(pp+1,pp+n+1,kmp);
        len=0;
        for(int i=1,j=Max; i<=n; i++)
        {
            if(pp[i].bj==j) pp[i].bj=len;
            else
            {
                j=pp[i].bj;
                pp[i].bj=++len;
            }
        }
        printf("%d\n",len);
        memset(vi,1,sizeof(vi));
        for(int i=1;i<=n;i++)
            if(pp[i].bj!=pp[i-1].bj) bfs(i);
    }
    return 0;
}