题干：

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. 

Find the both smallest amount of time it will take Bessie to join her cow friends. 

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

 

解题报告：

      很坑的一道单源最短路。Dijkstra或Spfa均可解。思想和这题类似【HDU - 1546】 Idiomatic Phrases Game（Dijkstra）

AC代码：

#include <iostream>
#include<cstring>
#include<cmath> 
#include<cstdio>
#include<queue> 
#include <cfloat>
using namespace std;int r,c;
double v;
const double INF = 0x3f3f3f3f; 
const double eps = 1e-8;
double dis[105][105];
int maze[105][105];
bool vis[105][105];//此题以二维数组表示一个点，就不对每个点打标记了。。。比如3*3的矩阵，打标记就是点1~9，此题不这样做了，，不然太麻烦。 
int head[105];
int nx[4] = {0,1,0,-1};
int ny[4] = {1,0,-1,0};
//即，此题把一维表示点，都扩展成二维即可。 此题用二维表示点 
struct Point {int x,y;double t;bool operator<(const Point & b) const {return t>b.t; }Point(){}Point(int x,int y,double t):x(x),y(y),t(t) {}
}p;
//默认从(1,1)跑到(r,c); 
void Dijkstra() {dis[1][1] = 0;
//	int all = r*c;//共r*c个点 double minw = INF;priority_queue<Point > pq;pq.push(Point(1,1,0));while(!pq.empty()) {int nxx,nyy;//找点 Point cur = pq.top();pq.pop();if(vis[cur.x][cur.y] == 1) continue;vis[cur.x][cur.y] = 1;if(cur.x == r && cur.y == c) break;for(int k = 0; k<4; k++) {nxx = cur.x+nx[k];nyy = cur.y+ny[k];if(vis[nxx][nyy] == 1) continue;if(nxx<1 || nxx>r || nyy<1 || nyy>c) continue;double t = cur.t + 1.0 / ( pow(2,maze[1][1]-maze[cur.x][cur.y])*v );if(dis[nxx][nyy] > t) {dis[nxx][nyy] = t;pq.push(Point(nxx,nyy,dis[nxx][nyy]));}
//			dis[nxx][nyy] = dis[nxx][nyy] == INF ? t : min(t,dis[nxx][nyy]);	
//			pq.push(Point(nxx,nyy,dis[nxx][nyy]));			}}printf("%.2f\n",dis[r][c]);
} void init() {memset(vis,0,sizeof(vis));
//	memset(dis,INF,sizeof(dis));for(int i = 1; i<=r; i++) {for(int j = 1; j<=c; j++) {dis[i][j] = DBL_MAX;}}memset(maze,0,sizeof(maze));
}
int main()
{scanf("%lf%d%d",&v,&r,&c); init();for(int i = 1; i<=r; i++) {for(int j = 1; j<=c; j++) {scanf("%d",&maze[i][j]);}}Dijkstra() ;return 0 ;
}

总结：