zoj 3652 Maze
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.
Author: ZHANG, Ruijie
这道题感觉特别恶心，题超长，翻译看哭我。。。而且限制条件有点烦，注意看懂题意才能写。。
题意：
题目大意：打怪兽。给一个n*m的地图，其中1~5表示该位置被第i个怪兽管理着，0表示空地可以走，-1表示这个位置不可以走。
再输入一个k，接下去k行怪兽所在的坐标。
最后一行就是Celica的起点和终点，问最少需要多少回合才能走到终点。
特别注意：
1、每一个回合开始，Celica的初始体力值为L，每移动一次，体力值减1，当体力值==0时，开始下一回合。
2、只要进入怪兽所管理的位置，体力值就瞬间变为0。
3、如果进入到怪兽所在的位置，就要将怪兽打死，体力值变为0,。与此同时，地图上怪兽的位置和怪兽所控制的位置全部变为0（无障碍空地）。
4、起点没有怪，但可能被某一个怪兽控制。
5、如果起点和终点重合，直接输出0。即回合数为0.
6、怪兽不一定在他控制的位置，被一号怪兽控制的区域也可以是2号怪兽。
7、复活过得地方不会在死一次了，并且起始点就算是复活地。
8、在到达终点的时候，体力值刚好变为0的话，不会再另外算一个回合，需要特殊判断。
9、如果不能到达终点的话，记得输出We need God's help!
附上代码：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstring> 4 #include <queue> 5 using namespace std; 6 int map[55][55],m,n,w; 7 int f[4][2]= {0,1,0,-1,1,0,-1,0}; 8 int k,a[10],b[10]; 9 int x1,y1,x2,y2; 10 int vis[55][55][100]; //使用状态压缩，记录怪物被杀的情况 11 12 struct node 13 { 14 int x,y; //当前的坐标 15 int step; //当前的体力 16 int t; //回合数 17 int flag; //标记怪物的存活情况 18 friend bool operator < (node n1,node n2) //优先队列，回合少的优先，回合相等，剩余体力多的优先 19 { 20 if(n1.t!=n2.t) 21 return n1.t>n2.t; 22 return n1.step<n2.step; 23 } 24 } s1,s2; 25 26 void BFS() 27 { 28 priority_queue<node> q; 29 while(!q.empty()) 30 q.pop(); 31 memset(vis,0,sizeof(vis)); 32 s1.x=x1; 33 s1.y=y1; 34 s1.step=w; //初始体力为w 35 s1.t=1; 36 s1.flag=0; //初始为0，表示怪物全部存活 37 vis[x1][y1][0]=1; //记录这种状态已出现过 38 q.push(s1); 39 while(!q.empty()) 40 { 41 s1=q.top(); 42 q.pop(); 43 if(s1.x==x2&&s1.y==y2) //走到了终点 44 { 45 if(s1.step==w) //如果此时体力为满，则说明上一个回合体力为0时，刚好到达终点，因此当前回合数减一 46 printf("%d\n",s1.t-1); 47 else 48 printf("%d\n",s1.t); 49 return; 50 } 51 for(int i=0; i<4; i++) 52 { 53 s2.x=s1.x+f[i][0]; 54 s2.y=s1.y+f[i][1]; 55 s2.flag=s1.flag; 56 s2.t=s1.t; 57 if(s2.x>=1&&s2.x<=n&&s2.y>=1&&s2.y<=m&&map[s2.x][s2.y]!=-1&&!vis[s2.x][s2.y][s2.flag]) 58 //不能超出地图范围，不能是墙，这种状态不能出现过 59 { 60 vis[s2.x][s2.y][s2.flag]=1; 61 if(map[s2.x][s2.y]==0||(s2.flag&(1<<(map[s2.x][s2.y])))!=0) //判断怪物是否被打死 62 s2.step=s1.step-1; //打死了或没怪物则体力减一 63 else 64 s2.step=0; //否则体力清零 65 for(int j=1; j<=k; j++) 66 if(s2.x==a[j]&&s2.y==b[j]) //如果走到有怪物的地方，杀死怪物，用flag记录 67 { 68 s2.flag=s2.flag|(1<<j); 69 } 70 if(s2.step==0) //如果体力为0，则开启下一回合，并且体力恢复到w 71 { 72 s2.t++; 73 s2.step=w; 74 } 75 q.push(s2); // 入队 76 } 77 78 } 79 } 80 printf("We need God's help!\n"); //如果搜不到，则输出这个 81 } 82 83 int main() 84 { 85 int i,j; 86 while(~scanf("%d%d%d",&n,&m,&w)) 87 { 88 for(i=1; i<=n; i++) 89 for(j=1; j<=m; j++) 90 scanf("%d",&map[i][j]); 91 scanf("%d",&k); //输入怪物个数 92 for(i=1; i<=k; i++) 93 scanf("%d%d",&a[i],&b[i]); //记录每个怪物的位置 94 scanf("%d%d%d%d",&x1,&y1,&x2,&y2); //输入起点和终点 95 if(x1==x2&&y1==y2) //若起点就是终点，则回合数为0 96 { 97 printf("0\n"); 98 continue; 99 } 100 BFS(); 101 } 102 return 0; 103 }
