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最新推荐文章于 2021-10-10 20:25:35 发布

翻译最新推荐文章于 2021-10-10 20:25:35 发布 · 343 阅读

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#lift #button #go #input #up #output

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本文探讨了一种特殊的电梯系统，该系统在每一层楼都有特定的上下行移动距离。通过输入建筑层数、起始楼层、目标楼层以及每层楼的移动参数，我们介绍了如何使用算法计算从起始楼层到达目标楼层所需的最少按钮按下次数。包括使用迪杰斯特拉算法和广度优先搜索两种方法来解决路径优化问题。

A strange lift

Time 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.
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"?

Input

The input consists of several test cases.,Each test case contains two lines.
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.
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
3 3 1 2 5
0

Sample Output

Recommend

8600

#include<stdio.h>
#include<memory.h>
#define inf 100000000
#define size 201
int map[size][size];

int dist[size];
int visit[size];

int min;
int dir;

int a;
int b;
int n;

int x;

void dijkstra(void)
{
    for(int i=1;i<=n;i++)
        dist[i]=map[a][i];
    
    visit[a]=1;
    dist[a]=0;
    for(int i=1;i<=n;i++){
        min=inf;
        for(int j=1;j<=n;j++)
            if(dist[j]<min&&!visit[j]){
                min=dist[j];
                dir=j;
            }
        if(min==inf)
            break;
        visit[dir]=1;
        for(int j=1;j<=n;j++)
            if(!visit[j]&&dist[j]>dist[dir]+map[dir][j])
                dist[j]=dist[dir]+map[dir][j];
    }
}
int main(void) {
    while (scanf("%d", &n) &&n) {
        memset(visit,0,sizeof(visit));
        scanf("%d %d", &a, &b);
        for (int i = 1; i <= n; i++) {
            for (int j = 1; j <= n; j++)
                map[i][j] = inf;
            scanf("%d", &x);
            if (i + x <= n)
                map[i][i + x] = 1;
            if (i - x >= 1)
                map[i][i - x] = 1;
        }
		dijkstra();
        if (dist[b] == inf)
            printf("-1\n");
        else
            printf("%d\n", dist[b]);
    }
    return 0;
}

bfs:

#include<iostream>
#include<cstring>
using namespace std;

typedef struct {
    int floor;
    int time;
} Q;
int go[201];
int visit[201];
Q queue[10000];

int bfs(int n, int a, int b) {
    memset(visit, 0, sizeof (visit));
    memset(queue, 0, sizeof (queue));
    int qhead = 0;
    int qtail = 0;
    queue[qtail].floor = a;
    queue[qtail].time = 0;
    visit[queue[qhead].floor] = 1;
    qtail++;
    while (qhead < qtail) {
        if ((queue[qhead].floor + go[queue[qhead].floor] >= 1)
                && (queue[qhead].floor + go[queue[qhead].floor] <= n)
                && !visit[queue[qhead].floor + go[queue[qhead].floor]]) {
            if (queue[qhead].floor + go[queue[qhead].floor] == b)
                return queue[qhead].time + 1;
            visit[queue[qhead].floor + go[queue[qhead].floor]] = 1;
            Q tmp;
            tmp.floor = queue[qhead].floor + go[queue[qhead].floor];
            tmp.time = queue[qhead].time + 1;
            queue[qtail++] = tmp;
        }
        if ((queue[qhead].floor - go[queue[qhead].floor] >= 1)
                && (queue[qhead].floor - go[queue[qhead].floor] <= n)
                && !visit[queue[qhead].floor - go[queue[qhead].floor]]) {
            if (queue[qhead].floor - go[queue[qhead].floor] == b)
                return queue[qhead].time + 1;
            visit[queue[qhead].floor - go[queue[qhead].floor]] = 1;
            Q tmp;
            tmp.floor = queue[qhead].floor - go[queue[qhead].floor];
            tmp.time = queue[qhead].time + 1;
            queue[qtail++] = tmp;
        }
        qhead++;
    }
    return -1;
}

int main(void) {
    int n, a, b;
    while (cin >> n, n) {
        cin >> a >> b;
        for (int i = 1; i <= n; i++)
            cin >> go[i];
        if (a == b)
            cout << '0' << endl;
        else
            cout << bfs(n, a, b) << endl;
    }
    return 0;
}