POJ 2253 Frogger 图论 Dijkstra算法

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Frogger Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 21864   Accepted: 7122 Description Freddy Frog is sitting on a stone in the middle of a lake. Suddenly he notices Fiona Frog who is sitting on another stone. He plans to visit her, but since the water is dirty and full of tourists' sunscreen, he wants to avoid swimming and instead reach her by jumping. Unfortunately Fiona's stone is out of his jump range. Therefore Freddy considers to use other stones as intermediate stops and reach her by a sequence of several small jumps. To execute a given sequence of jumps, a frog's jump range obviously must be at least as long as the longest jump occuring in the sequence. The frog distance (humans also call it minimax distance) between two stones therefore is defined as the minimum necessary jump range over all possible paths between the two stones. You are given the coordinates of Freddy's stone, Fiona's stone and all other stones in the lake. Your job is to compute the frog distance between Freddy's and Fiona's stone. Input The input will contain one or more test cases. The first line of each test case will contain the number of stones n (2<=n<=200). The next n lines each contain two integers xi,yi (0 <= xi,yi <= 1000) representing the coordinates of stone #i. Stone #1 is Freddy's stone, stone #2 is Fiona's stone, the other n-2 stones are unoccupied. There's a blank line following each test case. Input is terminated by a value of zero (0) for n. Output For each test case, print a line saying "Scenario #x" and a line saying "Frog Distance = y" where x is replaced by the test case number (they are numbered from 1) and y is replaced by the appropriate real number, printed to three decimals. Put a blank line after each test case, even after the last one. Sample Input 2 0 0 3 4 3 17 4 19 4 18 5 0 Sample Output Scenario #1 Frog Distance = 5.000 Scenario #2 Frog Distance = 1.414 这道题又是一个Dijkstra算法,这个算法一看就不容易,毕竟这个单词主页君也不知道肿么读,但是这个题运用这个模板还是很容易的,首先,建立这个图,其实这个图很容易建立,就是两点间的距离,这样运用勾股定理就可以得到,然后,带入Dijkstra算法,求出0点到1点的最短路径,从0点开始,只不过把原来0到i最短路径判断改为从0到i可以每次跳跃的最近线段,然后输出d[1]就可以AC了。。。还是灰常简单的嘛!!! 下面是AC代码: #include<cstdio> #include<cmath> #include<iostream> using namespace std; struct st { int x,y; }point[500]; double G[500][500]; double max(double a,double b) { if(a>b) return a; else return b; } double Dijkstra(int n) { int i,j,w,mark[205]; double minc,d[205]; for (i=0;i<n;i++) mark[i]=0; for (i=0;i<n;i++) { d[i]=G[0][i]; } mark[0]=1;d[0]=0; for (i=1;i<n;i++) { minc=100000000; w=0; for (j=0;j<n;j++) if ((mark[j]==0)&&(minc>=d[j])) {minc=d[j];w=j;} mark[w]=1; for (j=0;j<n;j++) if ((mark[j]==0)&&(d[j]>max(d[w],G[w][j]))) d[j]=max(d[w],G[w][j]); } return d[1]; } int main() { int i,j,t=1,n; double item; while(1) { scanf("%d",&n); if(n==0) break; for(i=0;i<n;i++) scanf("%d%d",&point[i].x,&point[i].y); for(i=0;i<n;i++) for(j=0;j<n;j++) if(i!=j) { G[i][j]=sqrt((point[j].x-point[i].x)*(point[j].x-point[i].x)+(point[j].y-point[i].y)*(point[j].y-point[i].y)); G[j][i]=G[i][j]; } item=Dijkstra(n); printf("Scenario #%d\n",t); printf("Frog Distance = %.3f",item); cout<<endl<<endl; t++; } return 0; }

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