标签:style less temp add sele real lan type from
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.
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.
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.
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
4
0 11
10 21
10 35
4 13
-12 12
12 12
-12 -12
12 -12
0
我的答案(最大N没过)
1 #include <stdio.h> 2 #include <stdlib.h> 3 #include <unistd.h> 4 #include <math.h> 5 6 #define QMAXSIZE 10000 7 8 struct Crocodile { 9 int x; 10 int y; 11 int Visited; 12 int Path; 13 }; 14 typedef struct Crocodile *Point; 15 16 //队列部分 17 struct QNode { 18 int Data[QMAXSIZE]; 19 int rear; 20 int front; 21 }; 22 typedef struct QNode *Queue; 23 24 int IsEmpty(Queue Q) 25 { 26 return (Q->rear == Q->front); 27 } 28 29 void AddQ(Queue PtrQ, int item) 30 { 31 if((PtrQ->rear+1)%QMAXSIZE == PtrQ->front) { 32 printf("Queue full"); 33 return; 34 } 35 PtrQ->rear = (PtrQ->rear+1)%QMAXSIZE; 36 PtrQ->Data[PtrQ->rear] = item; 37 } 38 39 int DeleteQ(Queue PtrQ) 40 { 41 if(PtrQ->front == PtrQ->rear) { 42 printf("Queue empty"); 43 return -1; 44 } else { 45 PtrQ->front = (PtrQ->front+1)%QMAXSIZE; 46 return PtrQ->Data[PtrQ->front]; 47 } 48 } 49 50 void PrintQ(Queue PtrQ) 51 { 52 int i; 53 printf("[Queue]: "); 54 for(i=(PtrQ->front+1)%QMAXSIZE;i!=(PtrQ->rear+1)%QMAXSIZE 55 ;i=(i+1)%QMAXSIZE) 56 printf("%d ", PtrQ->Data[i]); 57 printf("\n"); 58 }//end Queue 59 60 void ReadPoint(Point P, int N) 61 { 62 int i; 63 P[0].x = 0; 64 P[0].y = 0; 65 P[0].Visited = 0; 66 P[0].Path = -1; 67 for(i=1;i<N;i++) { //N=N+1 68 scanf("%d %d\n", &P[i].x, &P[i].y); 69 P[i].Visited = 0; 70 } 71 } 72 73 void PrintfPoint(Point P, int N) 74 { 75 int i; 76 for(i=0;i<N;i++) { //N=N+1 77 printf("P[%d] X:%d Y:%d\n", i, P[i].x, P[i].y); 78 } 79 printf("----------------------------\n"); 80 } 81 82 void PrintPath(Point P, int stand) 83 { 84 int Path[100], i; 85 // printf("[PrintPath] stand=%d\n", stand); 86 // if(P[stand].Path != -1) { 87 // PrintPath(P, P[stand].Path); 88 // printf("%d %d\n", P[stand].x, P[stand].y); 89 // } 90 if(stand==0) printf("1\n"); 91 else { 92 for(i=0;P[stand].Path != -1;i++) { 93 Path[i] = stand; 94 stand = P[stand].Path; 95 } 96 // printf("[PrintPaht] i=%d\n", i); 97 printf("%d\n", i+1); 98 for(i--;i>=0;i--) { 99 printf("%d %d\n", P[Path[i]].x, P[Path[i]].y); 100 } 101 } 102 } 103 104 double PointDistance(Point P1, Point P2) 105 { 106 return sqrt(pow((P1->x - P2->x), 2) + pow((P1->y - P2->y), 2)); 107 } 108 109 double FindMinPath(Point P, int stand) 110 { 111 while(P[stand].Path != 0) { 112 stand = P[stand].Path; 113 } 114 return PointDistance(&P[0], &P[stand]); 115 } 116 117 int IsUp(Point P, int stand, double D, int island) 118 { 119 int xlen = 50-abs(P[stand].x); 120 int ylen = 50-abs(P[stand].y); 121 if(island == 1 && (xlen<=(D+7.5) || ylen<=(D+7.5))) 122 return 1; 123 else if(stand!=1 && (xlen<=D || ylen <=D)) 124 return 1; 125 return 0; 126 } 127 128 int IsUseless(Point P, int stand) 129 { 130 if(abs(P[stand].x) <= 7.5 && abs(P[stand].y) <= 7.5 ) 131 return 1; 132 else if(abs(P[stand].x == 50 && abs(P[stand].y == 50))) 133 return 1; 134 else 135 return 0; 136 } 137 138 void Visit(Point P, int stand) 139 { 140 printf("[Point] P.x:%d y:%d\n", P[stand].x, P[stand].y); 141 } 142 143 int BFS(Point P, int N, double D, int stand) 144 { 145 Queue Q; 146 int S, i, endP=-1; 147 double dist, island = 0, minDist=100; 148 149 Q = (Queue)malloc(sizeof(struct QNode)*N); 150 // Visit(P, stand); //访问P[0] 151 P[stand].Visited = 1; 152 AddQ(Q, stand); 153 154 while(!IsEmpty(Q)) { 155 // PrintQ(Q); 156 S = DeleteQ(Q); //提取队列 157 if(S == 0) //是否在岛上 158 island = 1; 159 else 160 island = 0; 161 if(IsUp(P, S, D, island)) { 162 endP = S; 163 break; 164 } 165 166 for(i=1;i<N;i++) { 167 if(IsUseless(P, i)) continue; 168 dist = PointDistance(&P[S], &P[i]); 169 if(!P[i].Visited && dist<=(D+(double)island*7.5)) { //未被访问且能跳到 170 // Visit(P, i); 171 P[i].Path = S; 172 if(IsUp(P, i, D, island)) { 173 // printf("[Ok] Here Point can go up\n"); 174 // printf("[Path] \n"); 175 // PrintPath(P, i); 176 // printf("\n"); 177 double temp; 178 temp = FindMinPath(P, i); 179 if(minDist > temp) { 180 minDist = temp; 181 endP = i; 182 } 183 } 184 P[i].Visited = 1; 185 AddQ(Q, i); 186 // dist = D+1; 187 } 188 } 189 } 190 return endP; 191 } 192 193 int main() 194 { 195 int N, endP; 196 double D; 197 Point P; 198 scanf("%d %lf", &N, &D); 199 N++; //从1开始 200 P = (Point)malloc(sizeof(struct Crocodile)*N); 201 ReadPoint(P, N); 202 // PrintfPoint(P, N); 203 204 endP = BFS(P, N, D, 0); 205 // printf("minP:%d\n", endP); 206 if(endP == -1) 207 printf("0\n"); 208 else { 209 PrintPath(P, endP); 210 } 211 return 0; 212 }
07-图5 Saving James Bond - Hard Version(30 分)
标签:style less temp add sele real lan type from
原文地址:https://www.cnblogs.com/ch122633/p/8994750.html