POJ【2253】Frogger（最小最大值）

题目链接：http://poj.org/problem?id=2253

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<cmath>
using namespace std;
#define INF 0x3f3f3f3f
const int N = 200 + 10;
int n;
double a[N][N],d[N],inq[N];
struct Stone {
	int x,y;
} s[N];
void SPFA(int s)
{
	for(int i = 0; i < n; i++) {
		d[i] = INF;
		inq[i] = 0;
	}
	d[s] = 0;//从起点到终点的路径中某一权值最小的边
	queue<int> q;
	q.push(s);
	while(!q.empty()) {
		int u = q.front();q.pop();
		inq[u] = 0;
		for(int v = 0; v < n; v++)
			if(d[v] > max(d[u],a[u][v])) {//修改松弛操作
				//d[v] = min(d[v],max(d[u],a[u][v]));
				d[v] = max(d[u],a[u][v]);
				if(!inq[v]) {
					inq[v] = 1;
					q.push(v);
				}
			}
	}
}
double calc(Stone s1,Stone s2)
{
	return sqrt(1.0 * (s1.x - s2.x) * (s1.x - s2.x) + 1.0 * (s1.y - s2.y) * (s1.y - s2.y));
}
int main()
{
	int k = 0;
	while(scanf("%d",&n) && n) {
		for(int i = 0; i < n; i++)
			for(int j = 0; j < n; j++)
				a[i][j] = INF;
		for(int i = 0; i < n; i++)
			scanf("%d%d",&s[i].x,&s[i].y);
		for(int i = 0; i < n; i++)
			for(int j = i + 1; j < n; j++)
				a[i][j] = a[j][i] = calc(s[i],s[j]);
		SPFA(0);
		if(k++) printf("\n");
		printf("Scenario #%d\n",k);
		printf("Frog Distance = %.3f\n",d[1]);//注意printf不要使用lf
	}
	return 0;
}

标签：int,s2,s1,inq,++,POJ,include,2253,Frogger
来源： https://blog.csdn.net/qq_43498798/article/details/100109303