06-图2 Saving James Bond
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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 whether or not he can escape.
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, print in a line "Yes" if James can escape, or "No" if not.
Sample Input 1:
14 2025 -15-25 288 4929 15-35 -25 2827 -29-8 -28-20 -35-25 -20-13 29-30 15-35 4012 12
Sample Output 1:
Yes
Sample Input 2:
4 13-12 1212 12-12 -1212 -12
Sample Output 2:
No
用的DFS,开始没注意中间是个圆。。。。最后才发现是读题的问题。
#include<iostream>#include<cstdlib>#include<cmath>using namespace std;int n,m,flag[101];bool f=false;struct point{int x,y;}p[101];typedef struct graphnode *Graph;struct graphnode{int G[101][101];};void Insertedge(Graph g,int x,int y){g->G[x][y]=1;g->G[y][x]=1;}Graph Create(){cin>>n>>m;Graph g=(Graph)malloc(sizeof(struct graphnode));g->G[0][0]=0;p[0].x=0;p[0].y=0;for(int i=1;i<=n;i++){cin>>p[i].x>>p[i].y;for(int j=0;j<i;j++){if(sqrt((p[j].x-p[i].x)*(p[j].x-p[i].x)+(p[j].y-p[i].y)*(p[j].y-p[i].y)*1.0)<=m+7.5){if(j==0)Insertedge(g,i,j);elseif(sqrt((p[j].x-p[i].x)*(p[j].x-p[i].x)+(p[j].y-p[i].y)*(p[j].y-p[i].y)*1.0)<=m)Insertedge(g,i,j);}}}return g;}void dfs(Graph g,int x){if((50-abs(p[x].x))<=m||(50-abs(p[x].y))<=m){f=true;return;}if(f){return;}for(int i=0;i<101;i++){if(g->G[x][i]==1&&!flag[i]){flag[x]=1;dfs(g,i);flag[x]=0;}}}int main(){Graph g=Create();flag[0]=1;dfs(g,0);if(f){cout<<"Yes"<<endl;}else{cout<<"No"<<endl;}return 0;}
- 06-图2 Saving James Bond
- 06-图2 Saving James Bond
- 06-图2 Saving James Bond
- 06-图2 Saving James Bond
- 06-图2 Saving James Bond
- 06-图2 Saving James Bond
- 06-图2 Saving James Bond
- 06-图2 Saving James Bond
- 06-图2 Saving James Bond
- 06-图2 Saving James Bond
- 06-图2 Saving James Bond
- 06-图2 Saving James Bond
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