PAT Mooc datastructure 6-1

Saving James Bond - Hard Version

This time let us consider the situation in the movie "Live and Let Die" in which James Bond, the world's most famous spy, was captured by a group of drug dealers. He was sent to a small piece of land at the center of a lake filled with crocodiles. There he performed the most daring action to escape -- he jumped onto the head of the nearest crocodile! Before the animal realized what was happening, James jumped again onto the next big head... Finally he reached the bank before the last crocodile could bite him (actually the stunt man was caught by the big mouth and barely escaped with his extra thick boot).

Assume that the lake is a 100 by 100 square one. Assume that the center of the lake is at (0,0) and the northeast corner at (50,50). The central island is a disk centered at (0,0) with the diameter of 15. A number of crocodiles are in the lake at various positions. Given the coordinates of each crocodile and the distance that James could jump, you must tell him a shortest path to reach one of the banks. The length of a path is the number of jumps that James has to make.

Input Specification:

Each input file contains one test case. Each case starts with a line containing two positive integers N (<=100), the number of crocodiles, and D, the maximum distance that James could jump. Then N lines follow, each containing the (x, y) location of a crocodile. Note that no two crocodiles are staying at the same position.

Output Specification:

For each test case, if James can escape, output in one line the minimum number of jumps he must make. Then starting from the next line, output the position (x, y) of each crocodile on the path, each pair in one line, from the island to the bank. If it is impossible for James to escape that way, simply give him 0 as the number of jumps. If there are many shortest paths, just output the one with the minimum first jump, which is guaranteed to be unique.

Sample Input 1:

17 15
10 -21
10 21
-40 10
30 -50
20 40
35 10
0 -10
-25 22
40 -40
-30 30
-10 22
0 11
25 21
25 10
10 10
10 35
-30 10

Sample Output 1:

4
0 11
10 21
10 35

1.题目分析

这个题目就是典型的最短路径问题~

使用BFS对图进行层级遍历，只要在任一层发现可以跳出的话，那么就逆序打印出来全体路径。

至于这个层次的问题，其实很好实现，因为在BFS入队的过程中，是将一个节点的链接节点依此入队。在入队的时候，只要将此节点的步点数加1写入下一层中即可完成层数的累积。

同时，为了完成最后的打印工作，需要使用一个数组，完成类似于链表的工作。每次压入新的节点时，就要将新节点对应的父节点数组值设置为其父节点的index，这样在推出时逆序压入堆栈，再重新打印出来，即可完成逆序打印工作。

有两个小点值得注意，一个是如果007非常牛逼，直接就可以跳出来的情况需要特殊处理一下。

二是按照题目要求，如果存在相同步数的路径，需要选出第一跳最短的那一条路径。为了实现这一功能，我在将鳄鱼节点录入时，先按照距离原点的距离升序排序。这样，在进行BFS的时候，总是先从短距离向长距离遍历，保证了第一跳最短距离的哪条路径最先被发现。

2.伪码实现

P = 007StartPoint;

if (Could Get Out at the StartPoint)

return FirstGetOut;

EnQueue(P,Queue)

while(IsNotEmpty(Queue))

{

P = DeQueue(Queue);

Find All the Point NextP that jump from P

{

if(NextP is not reached)

{

Jumped(NextP) = Jumped(P) + 1

father(NextP) =   P

if(CouldGetOutFrom(NextP))

return NextP;

EnQueue(NextP, Queue)

}

}

}

return CouldNotJumpOut

3. 通过代码:

 #define NONE -1
 #define DISK 100000
 #define MISTAKE -11
 #define FIOUT 5000000

 #include <stdio.h>
 #include <stdlib.h>

 static int jumped[][];

 void InitialJumped()
 {
     int i;
     int j;
     for(i=;i<=;i++)
     {
         for(j=;j<=;j++)
         {
             jumped[i][j]=NONE;
         }
     }
 }

 int IsJumped(int x,int y)
 {
     if(jumped[x+][y+] != NONE)
     {
         return ;
     }
     else
     {
         return NONE;
     }
 }

 void JumpOn(int x,int y,int jumpNum)
 {
     jumped[x+][y+]= jumpNum;
 }

 int GetJump(int x,int y)
 {
     return jumped[x+][y+];
 }

 /////////End of Visited /////////////

 //////////Begin of Croc//////////////

 typedef struct Croc{
     //location of this Croc
     int x;
     int y;
     int dis;
 }tCroc;

 void SetDis(tCroc* C)
 {
     C->dis = (C->x)*(C->x) + (C->y)*(C->y);
 }

 int GetDis(tCroc* C)
 {
     return C->dis;
 }

 //If bond could reach from ori to des.
 int Reachable(tCroc* ori,tCroc* des,int step)
 {
     int oriX;
     int oriY;
     int desX;
     int desY;
     int distanceSquare;
     int stepSquare;

     oriX = ori->x;
     oriY = ori->y;

     desX = des->x;
     desY = des->y;

     distanceSquare = (oriX-desX)*(oriX-desX)+(oriY-desY)*(oriY-desY);
     stepSquare = step*step;

     if(stepSquare >= distanceSquare)
     {
         return ;
     }
     else
     {
         return ;
     }
 }

 int GetOut(tCroc *ori,int step)
 {
     if(ori->x + step >= )
     {
         return ;
     }
     if(ori->x - step <= -)
     {
         return ;
     }
     if(ori->y + step >= )
     {
         return ;
     }
     if(ori->y - step <= -)
     {
         return ;
     }
     return ;
 }

 //////////End of Croc////////////////

 /////////DFS of Croc/////////////////

 int DFSofCroc(tCroc *ori,tCroc list[],int numOfCroc,int step)
 {
     int i;
     int localStep;

     JumpOn(ori->x,ori->y,);

     if((ori->x == )&& (ori->y == ))
     {
         localStep = step+;
     }
     else
     {
         localStep = step;
     }

     if(GetOut(ori,localStep)==)
     {
         return ;
     }

     for(i = ;i<numOfCroc;i++)
     {
         if(Reachable(ori,&list[i],localStep)==)
         {
             if(IsJumped(list[i].x,list[i].y)==NONE)
             {
                 int result;
                 result =  DFSofCroc(&list[i],list,numOfCroc,step);
                 if(result == )
                 {
                     return ;
                 }
             }
         }
     }
     return ;
 }
 //////////////////End of DFS Croc without COUNT/////////////////

 /////////////////Begin of queue//////////////

 typedef struct queueNode{
     tCroc * thisCroc;
     struct queueNode * nextCroc;
 }QNode;

 typedef struct CrocQueue{
     QNode *head;
     QNode *tail;
 }tCrocQueue;

 tCrocQueue* InitialQueue()
 {
     tCrocQueue* temp = malloc(sizeof(tCrocQueue));
     temp->head = NULL;
     temp->tail = NULL;
     return temp;
 }

 void EnQueue(QNode *node,tCrocQueue *Q)
 {
     if(Q->head == NULL)
     {
         Q->head = node;
         Q->tail = node;
         return ;
     }
     else
     {
         Q->tail->nextCroc = node;
         Q->tail = node;
         return;
     }
 }

 QNode * DeQueue(tCrocQueue *Q)
 {
     if(Q->head == NULL)
     {
         return NULL;
     }
     else
     {
         QNode *temp = Q->head;
         if(Q->head == Q->tail)
         {
             Q->head = NULL;
             Q->tail = NULL;
         }
         else
         {
             Q->head = Q->head->nextCroc;
         }
         return temp;
     }
 }

 int IsQueueEmpty(tCrocQueue *Q)
 {
     if(Q->head == NULL)
     {
         return ;
     }
     else
     {
         return ;
     }
 }
 ////////////////End of queue /////////////////

 ////////////////Begin of BFS/////////////////

 int BFS(tCroc *ori, tCroc list[], int Sum, int step, tCrocQueue *Q, int father[])
 {
     int count;
     int myfather;
     myfather = DISK;
     count = ;
     JumpOn(ori->x,ori->y,count);
     if(GetOut(ori,step+)==)
     {
         return FIOUT;
     }
     QNode * thisNode = malloc(sizeof(QNode));
     thisNode->thisCroc = ori;
     EnQueue(thisNode,Q);
     while(IsQueueEmpty(Q)==)
     {
         int i ;
         int localStep;
         QNode* temp = DeQueue(Q);

         count = GetJump(temp->thisCroc->x,temp->thisCroc->y);

         if((temp->thisCroc->x == )&&(temp->thisCroc->y==))
         {
             localStep = step + ;
             myfather = DISK;
         }
         else
         {
             localStep = step;
             for(i = ; i < Sum ; i++)
             {
                 if((temp->thisCroc->x == list[i].x)&&(temp->thisCroc->y== list[i].y))
                 {
                     myfather = i;
                     break;
                 }
             }
         }

         for(i = ; i < Sum ;i++)
         {
             if(Reachable(temp->thisCroc,&list[i],localStep)==)
             {
                 if(IsJumped(list[i].x,list[i].y)==NONE)
                 {
                     JumpOn(list[i].x,list[i].y,(count+));
                     father[i] = myfather;

                     if(GetOut(&list[i],step)==)
                     {
                         return i;
                     }
                     thisNode = malloc(sizeof(QNode));
                     thisNode->thisCroc = &list[i];
                     EnQueue(thisNode,Q);
                 }
             }
         }
     }

     return NONE;
 }

 ////////////////End of BFS///////////////////
 ////////////////Begin Stack//////////////////
 typedef struct stackNode{
     tCroc *thisNode;
     struct stackNode * next;
 }tStackNode;

 typedef struct CrocStack{
     tStackNode* Top;
     tStackNode* Bottom;
 }tCrocStack;

 tCrocStack* InitialStack()
 {
     tCrocStack * temp = malloc(sizeof(tCrocStack));
     temp->Top = NULL;
     temp->Bottom = NULL;
     return temp;
 }

 void Push(tStackNode *node,tCrocStack * S)
 {
     if(S->Top == NULL)
     {
         S->Top = node;
         S->Bottom = node;
     }
     else
     {
         node->next = S->Top;
         S->Top = node;
     }
 }

 tStackNode *Pop(tCrocStack *S)
 {
     if(S->Top == NULL)
     {
         return NULL;
     }
     else
     {
         tStackNode * temp = S->Top;
         if(S->Top == S->Bottom)
         {
             S->Bottom = NULL;
             S->Top = NULL;
         }
         else
         {
             S->Top = S->Top->next;
         }
         return temp;
     }
 }

 int IsStackEmpty(tCrocStack *S)
 {
     if(S->Top == NULL)
     {
         return ;
     }
     else
     {
         return ;
     }
 }
 /////////////////End of Stack////////////////

 int main()
 {
     int numOfCrocs;
     int step;
     int i;

     InitialJumped();
     scanf("%d %d",&numOfCrocs,&step);

     tCroc crocs[numOfCrocs];
     int preCro[numOfCrocs];

     //Put all the cordinates X,Y into array
     for(i=;i<numOfCrocs;i++)
     {
         int j;
         int tempX;
         int tempY;
         scanf("%d %d",&tempX,&tempY);
         if( tempX >  || tempX< - || tempY >  || tempY < - )
         {
             i--;
             numOfCrocs--;
             continue;
         }
         crocs[i].x = tempX;
         crocs[i].y = tempY;
         SetDis(&crocs[i]);
         for(j=i-;j>=;j--)
         {
             if(crocs[j].x == crocs[i].x && crocs[j].y == crocs[i].y)
             {
                 i--;
                 numOfCrocs--;
                 j=MISTAKE;
                 break;
             }
         }
         if(j==MISTAKE)
         {
             continue;
         }

         for(j=i;j>;j--)
         {
             if(GetDis(&crocs[j])<GetDis(&crocs[j-]))
             {
                 tCroc temp = crocs[j-];
                 crocs[j-]=crocs[j];
                 crocs[j]=temp;
             }
             else
             {
                 break;
             }
         }
         preCro[i] = NONE;
     }

 //    printf("---------------\n");
 //    for(i=0;i<numOfCrocs;i++)
 //    {
 //        printf("%d %d %d\n",crocs[i].x,crocs[i].y,crocs[i].dis);
 //    }
 //    printf("---------------\n");

     tCroc *Zero = malloc(sizeof(tCroc));
     Zero->x = ;
     Zero->y = ;

     tCrocQueue * MyQueue = InitialQueue();

     int result;

     result = BFS(Zero,crocs,numOfCrocs,step,MyQueue,preCro);

     if(result == NONE)
     {
         printf("");
         return ;
     }
     else if(result == FIOUT)
     {
         printf("");
     }
     else
     {
         int Dis = GetJump(crocs[result].x,crocs[result].y);
         printf("%d\n",Dis+);
     }

     tCrocStack * MyStack = InitialStack();
     while(result != DISK)
     {
         tStackNode *temp = malloc(sizeof(tStackNode));
         temp->thisNode = &crocs[result];
         Push(temp,MyStack);
         result = preCro[result];
     }

     while(IsStackEmpty(MyStack)==)
     {
         int printX;
         int printY;
         tStackNode *temp = Pop(MyStack);

         printX = temp->thisNode->x;
         printY = temp->thisNode->y;
         printf("%d %d\n",printX,printY);
     }

 //    result = DFSofCroc(Zero,crocs,numOfCrocs,step);

 //    if(result == 1)
 //    {
 //        printf("Yes");
 //    }
 //    else
 //    {
 //        printf("No");
 //    }

     return ;
 }