搜索算法8之1014
2016-04-20 11:25
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1 题目编号:1014
2 题目内容:
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.<br>Here comes the problem: when you is on floor A,and you want to go to floor B,how many times at least he havt to press the button "UP" or "DOWN"?<br>
Input
The input consists of several test cases.,Each test case contains two lines.<br>The first line contains three integers N ,A,B( 1 <= N,A,B <= 200) which describe above,The second line consist N integers k1,k2,....kn.<br>A single 0 indicate the end of the input.
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<br>3 3 1 2 5<br>0
Sample Output
3
3 解题思路形成过程:有一个特别的电梯,第i层有一个对应的数字ki, 对于第i层按上升键up可升上到i+k[i]层,按下降键down到达i-k[i]层,到达的楼层最高不能超过n层,最低不能小于1层。给你一个起点A和终点B,问最少要按几次上升键或者下降键到达目的地。把每一层都看成一个节点,问题就可以变成求起点到终点的最短路径问题。用bfs广搜,只要作一下标记去过的地方,不能再去即可。
4 代码:#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <queue>
using namespace std;
int N, A, B;
int a[205];
bool map[205], flag;
struct node
{
int x, step;
}n1, n2, m;
int main()
{
int i;
while (cin>>N)
{
if (N == 0) break;
cin >> A >> B;
for (i = 1; i <= N; i++)
{
cin >> a[i];
map[i] = false;
}
flag = false;
n1.x = A; n1.step = 0;
queue<node> Q;
Q.push(n1);
map[n1.x] = true;
while (!Q.empty())
{
m = Q.front();
Q.pop();
if (m.x == B) //到达目标
{
flag = true;
break;
}
n1.x = m.x - a[m.x];
n2.x = m.x + a[m.x];
if (n1.x>0 && n1.x <= B && !map[n1.x]) //下去的
{
n1.step = m.step + 1;
map[n1.x] = true; //标记
Q.push(n1);
}
if (n2.x>0 && n2.x <= B && !map[n2.x]) //上去的
{
n2.step = m.step + 1;
map[n2.x] = true; //标记
Q.push(n2);
}
}
if (flag) cout<<m.step<<endl;
else cout << "-1"<< endl;
}
return 0;
}
2 题目内容:
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.<br>Here comes the problem: when you is on floor A,and you want to go to floor B,how many times at least he havt to press the button "UP" or "DOWN"?<br>
Input
The input consists of several test cases.,Each test case contains two lines.<br>The first line contains three integers N ,A,B( 1 <= N,A,B <= 200) which describe above,The second line consist N integers k1,k2,....kn.<br>A single 0 indicate the end of the input.
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<br>3 3 1 2 5<br>0
Sample Output
3
3 解题思路形成过程:有一个特别的电梯,第i层有一个对应的数字ki, 对于第i层按上升键up可升上到i+k[i]层,按下降键down到达i-k[i]层,到达的楼层最高不能超过n层,最低不能小于1层。给你一个起点A和终点B,问最少要按几次上升键或者下降键到达目的地。把每一层都看成一个节点,问题就可以变成求起点到终点的最短路径问题。用bfs广搜,只要作一下标记去过的地方,不能再去即可。
4 代码:#include <iostream>
#include <stdio.h>
#include <memory.h>
#include <queue>
using namespace std;
int N, A, B;
int a[205];
bool map[205], flag;
struct node
{
int x, step;
}n1, n2, m;
int main()
{
int i;
while (cin>>N)
{
if (N == 0) break;
cin >> A >> B;
for (i = 1; i <= N; i++)
{
cin >> a[i];
map[i] = false;
}
flag = false;
n1.x = A; n1.step = 0;
queue<node> Q;
Q.push(n1);
map[n1.x] = true;
while (!Q.empty())
{
m = Q.front();
Q.pop();
if (m.x == B) //到达目标
{
flag = true;
break;
}
n1.x = m.x - a[m.x];
n2.x = m.x + a[m.x];
if (n1.x>0 && n1.x <= B && !map[n1.x]) //下去的
{
n1.step = m.step + 1;
map[n1.x] = true; //标记
Q.push(n1);
}
if (n2.x>0 && n2.x <= B && !map[n2.x]) //上去的
{
n2.step = m.step + 1;
map[n2.x] = true; //标记
Q.push(n2);
}
}
if (flag) cout<<m.step<<endl;
else cout << "-1"<< endl;
}
return 0;
}
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