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B - Frogger——spfa()

2017-07-16 15:03 239 查看
Think:

1知识点:spfa()算法求最短路

2题意:n = 1 的青蛙要跳到 n = 2 的青蛙哪里,可以其它石头作为中介,询问在可达路径中青蛙至少要能跳多远

建议参考博客1

建议参考博客2

vjudge题目链接

以下为Accepted代码

#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>

using namespace std;

const int N = 240;
const int inf = 0x3f3f3f3f;

int n, x
, y
, vis
;
double e

, dis
;

void spfa();

int main(){
int k = 1;
while(scanf("%d", &n) && n){
memset(e, 0, sizeof(e));
for(int i = 1; i <= n; i++)
scanf("%d %d", &x[i], &y[i]);
for(int i = 1; i <= n; i++){
for(int j = i; j <= n; j++){
double x_2 = (double)((x[i]-x[j])*(x[i]-x[j]));
double y_2 = (double)((y[i]-y[j])*(y[i]-y[j]));
e[i][j] = e[j][i] = sqrt(x_2 + y_2);
}
}
spfa();
printf("Scenario #%d\n", k++);
printf("Frog Distance = %.3lf\n", dis[2]);
printf("\n");
}
return 0;
}
void spfa(){
queue <int> q;
for(int i = 1; i <= n; i++)
dis[i] = inf;
///memset(dis, inf, sizeof(dis));不能将double类型的数组初始化为无穷大
memset(vis, 0, sizeof(vis));
q.push(1);
dis[1] = 0, vis[1] = 1;
while(!q.empty()){
int t1 = q.front();
q.pop();
vis[t1] = 0;
for(int v = 1; v <= n; v++){
if(max(dis[t1], e[t1][v]) < dis[v]){
dis[v] = max(dis[t1], e[t1][v]);
if(vis[v] == 0){
q.push(v);
vis[v] = 1;
}
}
}
}
}
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标签:  spfa-算法 最短路