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恭喜福州大学杨楠获得[BestCoder Round #4]冠军（iPad Mini一部） 《BestCoder用户手册》下载 |
A strange liftTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Problem Description There is a strange lift.The lift can stop can at every floor as you want, and there is a number Ki(0 <= Ki <= N) on every floor.The lift have just two buttons: up and down.When you at floor i,if you press the button "UP" , you will go up Ki floor,i.e,you will go to the i+Ki th floor,as the same, if you press the button "DOWN" , you will go down Ki floor,i.e,you will go to the i-Ki th floor. Of course, the lift can‘t go up high than N,and can‘t go down lower than 1. For example, there is a buliding with 5 floors, and k1 = 3, k2 = 3,k3 = 1,k4 = 2, k5 = 5.Begining from the 1 st floor,you can press the button "UP", and you‘ll go up to the 4 th floor,and if you press the button "DOWN", the lift can‘t do it, because it can‘t go down to the -2 th floor,as you know ,the -2 th floor isn‘t exist. Input The input consists of several test cases.,Each test case contains two lines. Output For each case of the input output a interger, the least times you have to press the button when you on floor A,and you want to go to floor B.If you can‘t reach floor B,printf "-1". Sample Input 5 1 5 3 3 1 2 5 0 Sample Output 3 Recommend 8600 | We have carefully selected several similar problems for you: 1385 1242 1142 1217 2066 |
这道题刚拿到的时候，我就觉得和最短路径没关系。不过，在后来的思考中我慢慢发现了它与最短路径间的
联系。这道题要求解的是最少的按键次数并要求能够到达目的楼层，而且每个楼层只能向上或向下移动该楼层
规定的层数，那么从a楼层到规定楼层只需按一次键即可。这里，我们可以将这一过程模拟成最短路径中等的
从a地道另一个指定位置的过程，而两地的距离即权值就是1，那么上下楼层的问题可以看成是从一个地方
到另一个地方，而两地的间距都是1的移动过程。最后，在通过dijkstra算法来统计出从起点到终点所需的
按键次数即可，若可到达就输出次数，不可到达的话次数为无穷大，输出-1。
AC代码：
#include<stdio.h>
#include<string.h>
#define max 0x3f3f3f3f
int map[201][201];
int dist[201];
int floor[201];
void dijkstra(int num,int v)
{
bool vis[201];
memset(vis,0,sizeof(vis));
int i,j;
for(i=1;i<=num;i++)
{
dist[i]=map[v][i];
}
dist[v]=0;
vis[v]=1;
for(i=2;i<=num;i++)
{
int tmp=max;
int u=v;
for(j=1;j<=num;j++)
if((!vis[j])&&dist[j]<tmp)
{
u=j;
tmp=dist[j];
}
vis[u]=1;
for(j=1;j<=num;j++)
if((!vis[j])&&map[u][j]<max)
{
int newdist=dist[u]+map[u][j];
if(newdist<dist[j])
{
dist[j]=newdist;
}
}
}
}
int main()
{
int n;
while(scanf("%d",&n)!=EOF&&n)
{
memset(map,max,sizeof(map));
int start,end;
scanf("%d%d",&start,&end);
int i;
for(i=1;i<=n;i++)
{
scanf("%d",&floor[i]);
if(floor[i]+i<=n)
map[i][floor[i]+i]=1;
if(i-floor[i]>=1)
map[i][i-floor[i]]=1;
}
dijkstra(n,start);
if(dist[end]==max)
printf("-1\n");
else
printf("%d\n",dist[end]);
}
return 0;
}
hdu 1548 (dijkstra解法）（一次AC就是爽）,布布扣,bubuko.com