题目地址
https://pta.patest.cn/pta/test/16/exam/4/question/673
5-11 Saving James Bond - Hard Version (30分)
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 NN (le 100≤100), the number of crocodiles, and DD, the maximum distance that James could jump. Then NN lines follow, each containing the (x, y)(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)(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
Sample Input 2:
4 13
-12 12
12 12
-12 -12
12 -12
Sample Output 2:
0
/*
评测结果
时间 结果 得分 题目 编译器 用时（ms） 内存（MB） 用户
2017-07-04 18:58 答案正确 30 5-11 gcc 12 1
测试点结果
测试点 结果 得分/满分 用时（ms） 内存（MB）
测试点1 答案正确 15/15 1 1
测试点2 答案正确 3/3 2 1
测试点3 答案正确 2/2 2 1
测试点4 答案正确 3/3 1 1
测试点5 答案正确 6/6 2 1
测试点6 答案正确 1/1 12 1
解法，设一个空的起点，设一个空的终点
把第一跳能跳到的点和起点连起来
把最后能上岸的点和终点连起来
然后用dijkstra算法求起点到终点的最短路径
*/
#include<stdio.h>
#include<math.h>
#define MAX_CROCODILE 102
//set CROCODILE #101 to start Vertex,N+1 to finish Vertex
#define _START_IDX 101
#define _FINISH_IDX N+1
#define CENTRAL_ISLAND_R 7.5
#define BORDER 50
#define TRUE 1
#define FALSE 0
#define INF 10000
struct tCrocoVertex
{
int x;
int y;
int start;
int finish;
} gNodeTab[MAX_CROCODILE]; //主要是放坐标的，在HardVersion里面finish和start反而没什么用了
float gMatrix[MAX_CROCODILE][MAX_CROCODILE]; //放邻接矩阵
float gPathDist[MAX_CROCODILE];//放路径长度
float gFirstJump[MAX_CROCODILE];//放第一跳的长度，根据题意去重用
int gPathFrom[MAX_CROCODILE];//放最短路径的前一个节点
int gCollected[MAX_CROCODILE];//Dijkstra里面需要用的收纳表，看该点有没有被纳入集合内
int Min(int a,int b)
{
return a<b?a:b;
}
void InitTables()
{
int i;
for(i=0;i<MAX_CROCODILE;i++)
{
gPathDist[i]=INF;
gFirstJump[i]=INF;
gCollected[i]=0;
}
}
void InitMatrix()
{
int i,j;
for(i=0;i<= _START_IDX;i++)
for(j=0;j<=_START_IDX;j++)
gMatrix[i][j]=INF;
}
void FindStartAndFinishNode(int N,int jumpDistance)
{
int i,minFinishDist;
//minFinishDist放一个点到岸上的最短距离
float dist;
float startDist=jumpDistance+CENTRAL_ISLAND_R;
for(i=0;i<N;i++)
{
dist=sqrt((gNodeTab[i].x)*(gNodeTab[i].x)+(gNodeTab[i].y)*(gNodeTab[i].y));
if(dist<=startDist) //第一步跳出去够得着的点
{
gNodeTab[i].start=TRUE;
dist=dist-CENTRAL_ISLAND_R;
gFirstJump[i]=dist;
dist=1;//切换算距离还是算步数，算距离的话把这行注释掉
gMatrix[i][_START_IDX]=dist;
gMatrix[_START_IDX][i]=dist;
}
minFinishDist=abs(gNodeTab[i].x-BORDER);
minFinishDist=Min(abs(gNodeTab[i].x+BORDER),minFinishDist);
minFinishDist=Min(abs(gNodeTab[i].y-BORDER),minFinishDist);
minFinishDist=Min(abs(gNodeTab[i].x+BORDER),minFinishDist);
if(minFinishDist<=jumpDistance)
{
gNodeTab[i].finish=TRUE;
minFinishDist=1;//切换算距离还是算步数，算距离的话把这行注释掉
gMatrix[i][_FINISH_IDX-1]=minFinishDist;
gMatrix[_FINISH_IDX-1][i]=minFinishDist;
}
}
}
void BuildMatrix(int N,int jumpDistance)
{
int i,j;
float dist;
for(i=0;i<N;i++)
for(j=0;j<i;j++)
{
dist=sqrt( (gNodeTab[i].x-gNodeTab[j].x)*
(gNodeTab[i].x-gNodeTab[j].x)+
(gNodeTab[i].y-gNodeTab[j].y)*
(gNodeTab[i].y-gNodeTab[j].y));
if(dist<=jumpDistance)
{
dist=1;//切换算距离还是算步数，算距离的话把这行注释掉
gMatrix[i][j]=dist;
gMatrix[j][i]=dist;
}
}
}
int FindNextIDX(int N)//找出没有被收进集合的点中，dist最小的
{
int i,minIDX,minPath;
minIDX=-1;
minPath=INF;
for(i=0;i<_FINISH_IDX;i++)
{
if(gCollected[i]!=TRUE)
{
if(gPathDist[i]<minPath)
{
minPath=gPathDist[i];
minIDX=i;
}
}
}
return minIDX;
}
void Dijkstra(int N)
{
int i,nextIDX;
gPathDist[_START_IDX]=0;
nextIDX=_START_IDX;
while(nextIDX!=-1)
{
gCollected[nextIDX]=1;
for(i=0;i<_FINISH_IDX;i++)
{
if(gPathDist[i]>gPathDist[nextIDX]+gMatrix[nextIDX][i] ||
((int)gPathDist[i]==(int)(gPathDist[nextIDX]+gMatrix[nextIDX][i]) && gFirstJump[i]>gFirstJump[nextIDX] && (int)gPathDist[i]!= INF))
{
gPathDist[i]=gPathDist[nextIDX]+gMatrix[nextIDX][i];
gPathFrom[i]=nextIDX;
if(nextIDX!=_START_IDX)
gFirstJump[i]=gFirstJump[nextIDX];
}
}
nextIDX=FindNextIDX(N);
}
}
void DBG_ShowStatus(int N) //调试用函数
{
int i,j;
for(i=0;i<_FINISH_IDX;i++)
{
printf("ID:%d,x:%d,y:%d,start:%d,finish:%d
",i,gNodeTab[i].x,gNodeTab[i].y,gNodeTab[i].start,gNodeTab[i].finish);
}
for(i=0;i<_FINISH_IDX;i++)
{
for(j=0;j<_FINISH_IDX;j++)
printf("%4.1f ",gMatrix[i][j]);
printf("
");
}
for(i=0;i<_FINISH_IDX;i++)
printf("%d,%4.1f ,FJ:%.2f,%d
",i,gPathDist[i],gFirstJump[i],gPathFrom[i]);
}
void PrintPath(int i) //因为路径要正序输出，懒得用栈了，直接递归
{
if (i==_START_IDX)
return;
PrintPath(gPathFrom[i]);
printf("%d %d
",gNodeTab[i].x,gNodeTab[i].y);
}
void Output(int N,int D)
{
int i;
if(D>=50-CENTRAL_ISLAND_R)
{
printf("1");
return;
//备注，最后一个测试点卡这里了，如果跳跃距离足够远能直接跳上岸，那么直接打个1就行了。
}
if((int)gPathDist[_FINISH_IDX-1]==INF)
{
printf("0");
return;
}
printf("%d
",(int)gPathDist[_FINISH_IDX-1]);
PrintPath(gPathFrom[_FINISH_IDX-1]);
}
int main()
{
int i,N,D;
scanf("%d %d",&N,&D);
InitMatrix();
InitTables();
for(i=0;i<N;i++)
{
scanf("%d %d",&gNodeTab[i].x,&gNodeTab[i].y);
}
FindStartAndFinishNode(N,D);
BuildMatrix(N,D);
// DBG_ShowStatus(N);
Dijkstra(N);
Output(N,D);
}