hdu1072 bfs時間最佳化剪枝

標籤：

Nightmare

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

Problem DescriptionIgnatius 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. 

 

InputThe 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. 

 

OutputFor 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 Output4-113 挺簡單的bfs+dp，記憶化搜尋，只有當到dp[now]<dp[pre],才能到達改點，dp[now]表示當前到達這一點的剩餘時間聽ms bb集訓隊的時一下午，搞了題目都看了好久，回到寢室馬上就有思路了

#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\n");        for(int i=0;i<n;i++){            printf("%s\n",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;}/*33 32 1 11 1 01 1 34 82 1 1 0 1 1 1 01 0 4 1 1 0 4 11 0 0 0 0 0 0 11 1 1 4 1 1 1 35 81 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 */

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hdu1072 bfs時間最佳化剪枝