poj2253--Frogger(最短路径）解题报告

题目链接：http://poj.org/problem?id=2253

题意：求青蛙一到青蛙二之间经过所有跳跃点的最长路径中的最小值

dijkstra ac code：

#pragma GCC optimize(3,"Ofast","inline")
#pragma GCC optimize(2)
#include
#include
#include
#include
#include
#include
#define NIL -1
#define INF 0x3f3f3f3f

using namespace std;

const int maxn=210;
int n,vis[maxn];
double dis[maxn],graph[maxn][maxn];

struct node{
	double x,y;
};

node nd[maxn];

double distance(double x1,double x2,double y1,double y2)
{
	return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}

void dijkstra()
{
	for(int i=1;i<=n;i++){vis[i]=0;dis[i]=INF;}
	dis[1]=0;
	while(1){
		int minv=INF,u=-1;
		for(int i=1;i<=n;i++){if(!vis[i]&&dis[i]<minv){minv=dis[i];u=i;}}
		if(u==-1)break;
		vis[u]=1;
		for(int v=1;v<=n;v++){
			if(vis[v])continue;
			dis[v]=min(dis[v],max(dis[u],graph[u][v]));
		}
	}
}

int main()
{
	int kase=0;
	while(cin>>n){
		if(n==0)break;
		memset(graph,0,sizeof(graph));
		double x,y;
		for(int i=1;i<=n;i++){
			cin>>x>>y;
			nd[i].x=x;nd[i].y=y;
		}
		for(int i=1;i<=n;i++){
			for(int j=1;j<=i;j++){
				graph[i][j]=graph[j][i]=distance(nd[i].x,nd[j].x,nd[i].y,nd[j].y);//无向图记得把ij和ji位置的权值赋成同一个 
			}
		}
		dijkstra();
		printf("Scenario #%d\nFrog Distance = %.3lf\n\n",++kase,dis[2]);//输出格式也要小心，我之前一直在#后面加了个冒号，一直wa又没报pe，心好塞，后来才发现是输出格式的问题
	}
	return 0;
}

Floyd：