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 程式師世界 >> 編程語言 >> C語言 >> C++ >> 關於C++ >> ZOJ 3652Maze

ZOJ 3652Maze

編輯:關於C++

 

Maze

Time Limit : 4000/2000ms (Java/Other) Memory Limit : 131072/65536K (Java/Other)
Total Submission(s) : 44 Accepted Submission(s) : 13
Problem Description Celica is a brave person and believer of a God in the bright side. He always fights against the monsters that endanger humans. One day, he is asked to go through a maze to do a important task.

 

The maze is a rectangle of n*m, and Celica is at (x1, y1) at the beginning while he needs to go to (x2, y2). And Celica has a Mobility of l. When he moves a step the movility will decreased by 1. If his mobility equals to 0, he can't move anymore in this turn. And no matter how much mobility Celica uses in one turn, his movility will become l again in the next turn. And a step means move from one lattice to another lattice that has an adjacent edge.

However, due to the world's rule and the power of magic, there is something called Dominance in the maze. Some lattices may be dominated by some monsters and if Celica goes into these lattices, his mobility will be reduced to 0 at once by the monster's magic power. And monsters have strong Domain awareness so one lattice won't be dominated by more than one monster.

But luckily, Celica gets a strong power from his God so that he can kill this monsters easily. If Celica goes into a lattice which a monster stands on, he can kill the monster without anytime. If a monsters is killed, the lattices it dominates will no longer be dominated by anyone(or we can say they are dominated by Celica) and these lattices will obey the rule of mobility that normal lattices obey.

As for the task is so important that Celica wants to uses the least turn to go to (x2, y2). Please find out the answer.


PS1: It doesn't matter if Celica doesn't kill all the monsters in the maze because he can do it after the task and a monster may appear at a lattice that is not dominated by it, even a lattice that is not dominated by any monsters.

PS2: We define (1,1) as the top left corner. And monsters won't move.

PS3: No matter which lattice Celia gets in, the change of mobility happens first.

PS4: We promise that there is no two monsters have same position and no monster will appear at the start point of Celica.

Input


The first contains three integers, n, m, l.(1≤n, m≤50, 1≤l≤10)

Then there follows n lines and each line contains m integers. The j-th integer p in the line i describe the lattice in the i line and j row. If p euqals to -1, it means you can't get into it. If p euqals to 0, it means the lattice is not dominated by any monster. If p is larger than 0, it means it is dominated by the p-th monster.

And then in the n+2 line, there is an integer k(0≤k≤5) which means the number of monster.

Then there follows k lines. The i-th line has two integers mean the position of the i-th monster.

At last, in the n+k+3 lines, there is four integers x1, y1, x2, y2.

Output

If Celica can't get to the (x2, y2), output We need God's help!, or output the least turn Celica needs.

Sample Input

5 5 4
2 2 2 1 0
-1 2 2 -1 1
2 2 2 1 1
1 1 1 1 0
1 2 2 -1 0
2
4 2
1 1
5 1 1 5
5 5 4
1 1 1 1 1
1 2 2 -1 -1
2 2 -1 2 2
-1 -1 2 2 2
2 2 2 2 2
2
2 2
1 2
1 1 5 5

Sample Output

4
We need God's help!

Hit

In the first case, Celica goes to (4,1) in turn 1. Then he goes to (4,2) in turn 2. After he gets (4,2), kill the monster 1. He goes through (4,3)->(4,4)>(3,4)->(3,5) in turn 3. At last he goes (2,5)->(1,5) in turn 4.


 

狀態壓縮廣搜

有一個地圖,地圖劃分為幾個怪獸擁有的地盤,在殺死對應怪獸之前,在怪獸擁有的地盤上每回合只能移動一部,然後就進入下一回合。

每回合豬腳有l的行動力。求豬腳至少需要幾回合能達到目的地

 

因為怪物最多只有5個 所以可以使用狀態壓縮來保存隊列中每個點殺死怪獸的情況

四維數組標記判重即可

 

有幾個坑點

1、起點終點相同 1回合

2、到達終點時行動力為零也不需要等下一回合

3、怪獸不一定在自己的地盤上

 

 

#include
#include
#include
#include
#include
#include
#include
#include
#include
#include
#define inf 1<<30
#define LL long long
#define maxn 1<<24
using namespace std;
int Map[55][55];//地圖
int Mp[55][55];//怪獸所在點。。。有點懶了  
bool vis[55][55][1<<6][11];//標記
int dir[4][2]= {0,1,0,-1,1,0,-1,0};
int n,m,l;
int k;
struct node
{
    int x,y;
    int t;//移動力
    int T;//回合數
    int vis;//狀態壓縮
    friend bool operator < (node a,node b)
    {
        return a.T>b.T;
    }
    bool cheak()
    {
        if(x>0&&x<=n&&y>0&&y<=m)
            return true ;
        return false ;
    }
} st,e,ed;

void bfs()
{
    st.t=l;
    st.T=1;
    st.vis=0;
    priority_queueq;
    q.push(st);
    vis[st.x][st.y][st.vis][st.t]=true ;
    while(q.size())
    {
        st=q.top();
        q.pop();
        //cout<

 

 

 

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