hdu1072 bfs时间优化剪枝

Nightmare

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 9424    Accepted Submission(s): 4551

Problem Description

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

Sample Output

4

-1

13

 

---

挺简单的bfs+dp，记忆化搜索，只有当到dp[now]<dp[pre],才能到达改点，dp[now]表示当前到达这一点的剩余时间

听ms bb集训队的时一下午，搞了题目都看了好久，回到寝室马上就有思路了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <cmath>
#include <stack>
#include <queue>
#include <string>

const int inf = (1<<31)-1;
const int MAXN = 1e1;
using namespace std;

struct step{
    int x;
    int y;
    int t;
    int limit;
};

int mov[4][2]={-1,0,1,0,0,1,0,-1};
queue<step>q;
char G[MAXN][MAXN];
int dp[MAXN][MAXN];

int n,m;

int check(int x,int y){
    if(x<0||y<0||x>=n||y>=m)return 0;
    if(G[x][y]=='0')return 0;
    else return 1;
}

void init(){
    for(int i=0;i<n;i++){
        for(int j=0;j<m;j++)
            dp[i][j]=inf;
    }
}

int main()
{
    int t;
    int sx,sy;
    int flag;
    char ts[3];
    scanf("%d",&t);
    while(t--){
        flag = -1;
        scanf("%d%d",&n,&m);
        init();
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
               scanf("%s",ts);
               G[i][j] = ts[0];
                if(G[i][j]=='2'){
                    sx = i;
                    sy = j;
                }
            }
        }
      /*  printf("debug
");
        for(int i=0;i<n;i++){
            printf("%s
",G[i]);
        }*/
        step tp,fro;
        tp.x = sx;
        tp.y = sy;
        tp.t = 0;
        tp.limit = 0;
        q.push(tp);
        G[sx][sy]='1';
        dp[sx][sy]=0;
        int nx,ny;
        while(!q.empty()){

            fro = q.front();
            q.pop();
            if(G[fro.x][fro.y]=='3'){
                flag = fro.t;
                break;
            }
            if(fro.limit>=5)continue;
            for(int i=0;i<4;i++){
                nx = fro.x+mov[i][0];
                ny = fro.y+mov[i][1];
                if(check(nx,ny)){
                    if(G[nx][ny]=='4'){
                        tp.limit = 0;
                    }else{
                        tp.limit = fro.limit+1;
                    }
                    if(tp.limit<dp[nx][ny]){
                        tp.t = fro.t +1;
                        tp.x = nx;
                        tp.y = ny;
                        q.push(tp);
                        dp[nx][ny] = tp.limit;
                    }
                }
            }
        }
        while(!q.empty()){
            q.pop();
        }
       /*for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                cout<<dp[i][j]<<" ";
            }
            cout<<endl;
        }*/
        cout<<flag<<endl;
    }
    //cout << "Hello world!" << endl;
    return 0;
}

/*
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 
*/