HDU 1072 Nightmare(搜索-DFS)

Nightmare
Time Limit: 1000msMemory Limit: 32768KB This problem will be judged on HDU. Original ID: 1072
64-bit integer IO format: %I64d Java class name: Main
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Ignatius had a nightmare last night. He found himself in a labyrinth with a time bomb on him. The labyrinth has an exit, Ignatius should get out of the labyrinth before the bomb explodes. The initial exploding time of the bomb is set to 6 minutes. To prevent the bomb from exploding by shake, Ignatius had to move slowly, that is to move from one area to the nearest area(that is, if Ignatius stands on (x,y) now, he could only on (x+1,y), (x-1,y), (x,y+1), or (x,y-1) in the next minute) takes him 1 minute. Some area in the labyrinth contains a Bomb-Reset-Equipment. They could reset the exploding time to 6 minutes.

Given the layout of the labyrinth and Ignatius’ start position, please tell Ignatius whether he could get out of the labyrinth, if he could, output the minimum time that he has to use to find the exit of the labyrinth, else output -1.

Here are some rules:
1. We can assume the labyrinth is a 2 array.
2. Each minute, Ignatius could only get to one of the nearest area, and he should not walk out of the border, of course he could not walk on a wall, too.
3. If Ignatius get to the exit when the exploding time turns to 0, he can’t get out of the labyrinth.
4. If Ignatius get to the area which contains Bomb-Rest-Equipment when the exploding time turns to 0, he can’t use the equipment to reset the bomb.
5. A Bomb-Reset-Equipment can be used as many times as you wish, if it is needed, Ignatius can get to any areas in the labyrinth as many times as you wish.
6. The time to reset the exploding time can be ignore, in other words, if Ignatius get to an area which contain Bomb-Rest-Equipment, and the exploding time is larger than 0, the exploding time would be reset to 6.
Input
The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
Each test case starts with two integers N and M(1<=N,Mm=8) which indicate the size of the labyrinth. Then N lines follow, each line contains M integers. The array indicates the layout of the labyrinth.
There are five integers which indicate the different type of area in the labyrinth:
0: The area is a wall, Ignatius should not walk on it.
1: The area contains nothing, Ignatius can walk on it.
2: Ignatius’ start position, Ignatius starts his escape from this position.
3: The exit of the labyrinth, Ignatius’ target position.
4: The area contains a Bomb-Reset-Equipment, Ignatius can delay the exploding time by walking to these areas.
Output
For each test case, if Ignatius can get out of the labyrinth, you should output the minimum time he needs, else you should just output -1.
Sample Input
3
3 3
2 1 1
1 1 0
1 1 3
4 8
2 1 1 0 1 1 1 0
1 0 4 1 1 0 4 1
1 0 0 0 0 0 0 1
1 1 1 4 1 1 1 3
5 8
1 2 1 1 1 1 1 4
1 0 0 0 1 0 0 1
1 4 1 0 1 1 0 1
1 0 0 0 0 3 0 1
1 1 4 1 1 1 1 1

好的文章和知识被埋没是种罪过

这个bfs相对简单

bfs的思想就是遍历每一步，直到到终点
像圆一样，队列的头一定都是要走的下一步

看这个的bfs介绍，百度不到他写的感觉好浪费

代码
搜索的关键是要有搜索的思维和套路

#include<iostream>#include<cstring>#include<queue>#include<cstdio>using namespace std;struct node{    int x, y, time, step;}t, p, st;int n, m;int map[10][10];int dir[4][2] = { 0,1,0,-1,1,0,-1,0 };void bfs(){    queue<node> q;    q.push(st);    while (!q.empty())    {        p = q.front();        q.pop();        for (int i = 0; i < 4; i++)        {            t.x = p.x + dir[i][0];            t.y = p.y + dir[i][1];            t.step =p.step+ 1;            t.time =p.time- 1;            if (t.x > 0 && t.y > 0 && t.x <= n&&t.y <= m&&map[t.x][t.y] && t.time > 0)            {                if (map[t.x][t.y] == 4)                {                    //t.step + 1;                    t.time = 6;                    map[t.x][t.y] = 0;                }                if (map[t.x][t.y] == 3)                {                    cout << t.step << endl;                    return;                }                q.push(t);            }        }    }    puts("-1");    return;}int main(){    int t;    cin >> t;    while (t--)    {        int i, j;        cin >> n >> m;        for (i = 1; i <= n; i++)        {            for (j = 1; j <=m; j++)            {                scanf("%d", &map[i][j]);                if (map[i][j] == 2)                {                    st.x = i;                    st.y = j;                    st.time = 6;                    st.step = 0;                }            }        }        bfs();    }    return 0;}

他的

“`
#include
#include
#include
#define N 10
using namespace std;
int map[N][N],n,m;
int dir[4][2]=
{
{0,1}, /向右/
{0,-1}, /向左/
{-1,0}, /向下/
{1,0} /向上/
};
struct Node
{
int x,y;//记录坐标
int step,time;//步数和时间
}start;
void BFS()
{
queueq;//队列实现
Node q1,q2;//交换值，相当于temp
q.push(start);//将start放入队列

   //队列为空时说明 1.已经扫描到结果     //2.完全扫描结束（没找到结果）     while(!q.empty())     {        int i;        q1=q.front();//将队头的数据拿出来        q.pop();//将队头弹出        //开始搜索上下左右四个方向         for(i=0;i<4;i++)        {           q2.x=q1.x+dir[i][0];           q2.y=q1.y+dir[i][1];           q2.step=q1.step+1;           q2.time=q1.time-1;           //判断走这一步是否已经超出矩阵范围             //确定此步不是走过的(或墙)或者炸弹时间已到            if(q2.x>=0&&q2.y>=0&&q2.x<n&&q2.y<m&&map[q2.x][q2.y]!=0&&q2.time>0)           {              //说明找到答案，结束搜索               if(map[q2.x][q2.y]==3)              {                 printf("%d\n",q2.step);                 return;                                  }              else if(map[q2.x][q2.y]==4)              {//碰到时间调整器，可以恢复时间                    q2.time=6;                   map[q2.x][q2.y]=0; //标记已经走过                 }                q.push(q2);//将这一步放进队列                                          }        }      }       //队列扫完都没搜到答案，说明答案不存在     printf("-1\n");     return;  }  int main()  {      int i,j,T;      scanf("%d",&T);      while(T--)      {         scanf("%d %d",&n,&m);         for(i=0;i<n;i++)           for(j=0;j<m;j++)           {              scanf("%d",&map[i][j]);              if(map[i][j]==2)              {                 start.x=i;                 start.y=j;                 start.step=0;                 start.time=6;//时间初始化为6               }           }           BFS();       }      return 0;  }  

“`原链接

bfs

详解