poj 3037 -- Skiing spfa

Bessie and the rest of Farmer John's cows are taking a trip this winter to go skiing. One day Bessie finds herself at the top left corner of an R (1 <= R <= 100) by C (1 <= C <= 100) grid of elevations E (-25 <= E <= 25). In order to join FJ and the other cows at a discow party, she must get down to the bottom right corner as quickly as she can by travelling only north, south, east, and west.        

Bessie starts out travelling at a initial speed V (1 <= V <= 1,000,000). She has discovered a remarkable relationship between her speed and her elevation change. When Bessie moves from a location of height A to an adjacent location of eight B, her speed is multiplied by the number 2^(A-B). The time it takes Bessie to travel from a location to an adjacent location is the reciprocal of her speed when she is at the first location.   

Input

* Line 1: Three space-separated integers: V, R, and C, which respectively represent Bessie's initial velocity and the number of rows and columns in the grid.

* Lines 2..R+1: C integers representing the elevation E of the corresponding location on the grid.

Output

A single number value, printed to two exactly decimal places: the minimum amount of time that Bessie can take to reach the bottom right corner of the grid.

Sample Input

1 3 3
1 5 3
6 3 5
2 4 3

Sample Output

29.00

Hint

Bessie's best route is:
Start at 1,1 time 0 speed 1
East to 1,2 time 1 speed 1/16
South to 2,2 time 17 speed 1/4
South to 3,2 time 21 speed 1/8
East to 3,3 time 29 speed 1/4

题意是一头牛从左上角走到右下角，给你左上角的初始速度，每走一点，速度乘以2的两点的高度差次方，时间是速度的反比，问最短时间

这道题实在最短路的专题下挂出来的，可以看出从u到v的时间是由u的速度决定的，因为初始速度是固定的，而速度的的改变只和高度差有关，这样每一个点的速度都是固定的（这道题的关键），那么每一点到上下左右的时间都是确定得了，这就构成了图。

接下来就是如何找最短路了，这是我第一次用spfa算法写，spfa算法就是首先把源点加入队列里，然后对其相邻的边进行松弛，如果松弛成功，且这个点不在队列里，就把他加入队列里，写这个题的时候感觉和广搜十分相似，似乎这两个算法是同一个人写出来的，但有些地方还是不一样的，spfa里一个点可以多次加入队列，广搜只有一次

#include<stdio.h>#include<queue>#include<math.h>using namespace std;const double INF=10000000000;const int maxn=110;double v[maxn][maxn];double d[maxn][maxn];int h[maxn][maxn];bool vis[maxn][maxn];int R,C;double V;struct node{int x;int y;};int next1[4][2]={  {0,1},{1,0},{-1,0},{0,-1}};void spfa(){queue<node> q;struct node temp1,temp2;for(int i=1;i<=R;i++)   for(int j=1;j<=C;j++)   {      d[i][j]=INF;   vis[i][j]=false;      }temp1.x=1;temp1.y=1;q.push(temp1);vis[1][1]=true;d[1][1]=0;while(!q.empty()){temp1=q.front();q.pop();vis[temp1.x][temp1.y]=false;for(int i=0;i<=3;i++){temp2=temp1;temp2.x+=next1[i][0];temp2.y+=next1[i][1];if(temp2.x<1||temp2.y<1||temp2.x>R||temp2.y>C)   continue;double w=(1.0)/v[temp1.x][temp1.y];if(d[temp1.x][temp1.y]+w<d[temp2.x][temp2.y]){d[temp2.x][temp2.y]=d[temp1.x][temp1.y]+w;if(!vis[temp2.x][temp2.y]){    q.push(temp2);vis[temp2.x][temp2.y]=true;}}   }}  printf("%.2f\n",d[R][C]); }int main(void){scanf("%lf%d%d",&V,&R,&C);for(int i=1;i<=R;i++)    for(int j=1;j<=C;j++)    {    scanf("%d",&h[i][j]);    v[i][j]=V*pow(2.0,double(h[1][1])-double(h[i][j]));    }v[1][1]=V;        spfa();    return 0;}