hdu1072

                                                                                        Nightmare
                                  Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u


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

解题思想：

         逃生搜索的一种变形，主要在它的生存时间可以重置。题目大意是给你一个图，0代表墙，1代表可以走，2代表起始点，

3代表结束点，4代表时间重置点。对起始点进行BFS搜索，vis数组记录该点是否走过。

       在四个方向搜索时，如果满足坐标不越界，该点没走过并且该点不是墙，则进行if判断：（1）此时坐标恰好等于终点，返

回step；（2）maps等于1，时间减1；（3）maps等于4，时间重置，标记走过。如果最后队列为空时也不能走出，返回-1.

AC代码：

#include <functional>#include <algorithm>#include <iostream>#include <fstream>#include <sstream>#include <iomanip>#include <numeric>#include <cstring>#include <cassert>#include <cstdio>#include <string>#include <vector>#include <bitset>#include <queue>#include <stack>#include <cmath>#include <ctime>#include <list>#include <set>#include <map>#define REP0(i, n) for (int i=0;i<int(n);++i)#define REP1(i, n) for (int i=1;i<=int(n);++i)#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)using namespace std;#define REP(i, n) for (int i=0;i<int(n);++i)#define FOR(i, a, b) for (int i=int(a);i<int(b);++i)#define DWN(i, b, a) for (int i=int(b-1);i>=int(a);--i)#define REP_1(i, n) for (int i=1;i<=int(n);++i)#define FOR_1(i, a, b) for (int i=int(a);i<=int(b);++i)#define DWN_1(i, b, a) for (int i=int(b);i>=int(a);--i)#define REP_C(i, n) for (int n____=int(n),i=0;i<n____;++i)#define FOR_C(i, a, b) for (int b____=int(b),i=a;i<b____;++i)#define DWN_C(i, b, a) for (int a____=int(a),i=b-1;i>=a____;--i)#define REP_N(i, n) for (i=0;i<int(n);++i)#define FOR_N(i, a, b) for (i=int(a);i<int(b);++i)#define DWN_N(i, b, a) for (i=int(b-1);i>=int(a);--i)#define REP_1_C(i, n) for (int n____=int(n),i=1;i<=n____;++i)#define FOR_1_C(i, a, b) for (int b____=int(b),i=a;i<=b____;++i)#define DWN_1_C(i, b, a) for (int a____=int(a),i=b;i>=a____;--i)#define REP_1_N(i, n) for (i=1;i<=int(n);++i)#define FOR_1_N(i, a, b) for (i=int(a);i<=int(b);++i)#define DWN_1_N(i, b, a) for (i=int(b);i>=int(a);--i)#define REP_C_N(i, n) for (n____=int(n),i=0;i<n____;++i)#define FOR_C_N(i, a, b) for (b____=int(b),i=a;i<b____;++i)#define DWN_C_N(i, b, a) for (a____=int(a),i=b-1;i>=a____;--i)#define REP_1_C_N(i, n) for (n____=int(n),i=1;i<=n____;++i)#define FOR_1_C_N(i, a, b) for (b____=int(b),i=a;i<=b____;++i)#define DWN_1_C_N(i, b, a) for (a____=int(a),i=b;i>=a____;--i)//#define ECH(it, A) for (typeof(A.begin()) it=A.begin(); it != A.end(); ++it)#define ECH(it, A) for (__typeof(A.begin()) it=A.begin(); it != A.end(); ++it)#define REP_S(it, str) for (char*it=str;*it;++it) // 用于字符串的 .. .#define REP_G(it, u) for (int it=hd[u];it;it=suc[it]) // 用于图论的 .. .#define DO(n) for ( int ____n ## __line__ = n; ____n ## __line__ -- ; )#define REP_2(i, j, n, m) REP(i, n) REP(j, m)#define REP_2_1(i, j, n, m) REP_1(i, n) REP_1(j, m)#define REP_3(i, j, k, n, m, l) REP(i, n) REP(j, m) REP(k, l)#define REP_3_1(i, j, k, n, m, l) REP_1(i, n) REP_1(j, m) REP_1(k, l)#define ALL(A) A.begin(), A.end()#define LLA(A) A.rbegin(), A.rend()#define CPY(A, B) memcpy(A, B, sizeof(A))#define INS(A, P, B) A.insert(A.begin() + P, B)#define ERS(A, P) A.erase(A.begin() + P)#define BSC(A, X) find(ALL(A), X) // != A.end()#define CTN(T, x) (T.find(x) != T.end())#define SZ(A) int(A.size())#define PB push_back#define MP(A, B) make_pair(A, B)#define PTT pair<T, T>#define fi first#define se second#define ALL(A) A.begin(), A.end()#define LLA(A) A.rbegin(), A.rend()#define CPY(A, B) memcpy(A, B, sizeof(A))#define INS(A, P, B) A.insert(A.begin() + P, B)#define ERS(A, P) A.erase(A.begin() + P)#define BSC(A, X) find(ALL(A), X) // != A.end()#define CTN(T, x) (T.find(x) != T.end())#define SZ(A) int(A.size())#define PB push_back#define MP(A, B) make_pair(A, B)#define PTT pair<T, T>#define fi first#define se second#define Rush int T____; RD(T____); DO(T____)#define Rush int T____; RD(T____); DO(T____)#define Display(A, n, m) {                      \    REP(i, n){                                  \        REP(j, m) cout << A[i][j] << " ";       \        cout << endl;                         \    }                                           \}#define Display_1(A, n, m) {                    \    REP_1(i, n){                                \        REP_1(j, m) cout << A[i][j] << " ";     \        cout << endl;                         \    }                                           \}typedef long long LL;//typedef long double DB;typedef double DB;typedef unsigned UINT;typedef unsigned long long ULL;typedef vector<int> VI;typedef vector<char> VC;typedef vector<string> VS;typedef vector<LL> VL;typedef vector<DB> VD;typedef map<int, int> MII;typedef map<string, int> MSI;typedef map<LL, int> MLI;typedef map<DB, int> MDI;typedef pair<int, int> PII;typedef pair<int, bool> PIB;typedef pair<LL, LL> PLL;typedef vector<PII> VII;typedef vector<VI> VVI;typedef vector<VII> VVII;const int dx[] = {-1, 0, 1, 0};const int dy[] = {0, 1, 0, -1};const int MOD = 1000000007;const int INF = 0x3f3f3f3f;const LL INFF = 1LL << 60;const DB EPS = 1e-9;const DB OO = 1e15;const DB PI = acos(-1.0); //M_PI;#define N 81int n , m , i , j , k , t , flag;int maps[N][N] , vis[N][N] , d[4][2] = {-1, 0, 1, 0, 0, -1, 0 ,1};struct node{    int x , y , time ,step;};int bfs(node st , node ed){    memset(vis , 0 , sizeof(vis));    queue<node> q;    node now , next;    now.x = st.x;    now.y = st.y;    now.time = 6;    now.step = 0;    vis[st.x][st.y] = 1;    q.push(now);    while(!q.empty())    {        now = q.front();        q.pop();        REP(i , 4)        {            next.x = now.x + d[i][0];            next.y = now.y + d[i][1];            next.step = now.step + 1;            next.time = now.time - 1;            if(next.x >= 0 && next.x < n && next.y >= 0 && next.y < m && !vis[next.x][next.y] && next.time >= 1 && maps[next.x][next.y] != 0)            {                if(next.x == ed.x && next.y == ed.y)                    return next.step;                else if(maps[next.x][next.y] == 4)                {                    next.time = 6;                    vis[next.x][next.y] = 1;                }                else if(maps[next.x][next.y] == 1)                    next.time = now.time - 1;                q.push(next);            }        }    }    return -1;}int main(){    #ifdef ZH    freopen("in.txt","r",stdin);    #endif    int T;    scanf("%d",&T);    node st ,ed;    while(T--)    {        scanf("%d%d",&n,&m);        memset(maps , 0 , sizeof(maps));        REP(i , n)        {            REP(j , m)            {                scanf("%d",&maps[i][j]);                if(maps[i][j] == 2)                {                    st.x = i;                    st.y = j;                }                else if(maps[i][j] == 3)                {                    ed.x = i;                    ed.y = j;                }            }        }        printf("%d\n",bfs(st , ed));    }}