题干：
This story happened on the background of Star Trek.
Spock, the deputy captain of Starship Enterprise, fell into Klingon’s trick and was held as prisoner on their mother planet Qo’noS.
The captain of Enterprise, James T. Kirk, had to fly to Qo’noS to rescue his deputy. Fortunately, he stole a map of the maze where Spock was put in exactly.
The maze is a rectangle, which has n rows vertically and m columns horizontally, in another words, that it is divided into n*m locations. An ordered pair (Row No., Column No.) represents a location in the maze. Kirk moves from current location to next costs 1 second. And he is able to move to next location if and only if:
Next location is adjacent to current Kirk’s location on up or down or left or right(4 directions)
Open door is passable, but locked door is not.
Kirk cannot pass a wall
There are p types of doors which are locked by default. A key is only capable of opening the same type of doors. Kirk has to get the key before opening corresponding doors, which wastes little time.
Initial location of Kirk was (1, 1) while Spock was on location of (n, m). Your task is to help Kirk find Spock as soon as possible.
Input
The input contains many test cases.
Each test case consists of several lines. Three integers are in the first line, which represent n, m and p respectively (1<= n, m <=50, 0<= p <=10).
Only one integer k is listed in the second line, means the sum number of gates and walls, (0<= k <=500).
There are 5 integers in the following k lines, represents x i1, y i1, x i2, y i2, gi; when g i >=1, represents there is a gate of type gi between location (x i1, y i1) and (x i2, y i2); when g i = 0, represents there is a wall between location (x i1, yi1) and (x i2, y i2), ( | x i1 - x i2 | + | y i1 - y i2 |=1, 0<= g i <=p )
Following line is an integer S, represent the total number of keys in maze. (0<= S <=50).
There are three integers in the following S lines, represents x i1, y i1 and q irespectively. That means the key type of q i locates on location (x i1, y i1), (1<= qi<=p).
Output
Output the possible minimal second that Kirk could reach Spock.
If there is no possible plan, output -1.
Sample Input
4 4 9
9
1 2 1 3 2
1 2 2 2 0
2 1 2 2 0
2 1 3 1 0
2 3 3 3 0
2 4 3 4 1
3 2 3 3 0
3 3 4 3 0
4 3 4 4 0
2
2 1 2
4 2 1Sample Output
14题目大意：
给定一个棋盘，从(1,1)走到(n,m)有的任意两个格子之间的边视为：通路，门，墙。通路可以直接走，门必须早到相应的钥匙，墙永远不能通过。钥匙在一些给定点的格子中(同一个格子中可能有多把钥匙)，问采取怎样的走法可以得到最少的移动步数。
解题报告：
注意这题给的坐标代表的是一个房间，而不是点，读入的地图代表连接这两个房间的那条路是需要钥匙的还是不需要钥匙的还是墙。然后钥匙数状压一下，用邻接矩阵存图，然后用vis三维数组存当前状态和所得钥匙的情况数。vis[x][y][s]表示状态为s时到达(x,y)点是否已经到达过，s表示钥匙的得到情况的状态。
AC代码：
#include<bits/stdc++.h>
using namespace std;
int tx[4] = {0,1,0,-1};
int ty[4] = {1,0,-1,0};
int n,m;
int maze[55][55][55][55],key[55][55];
bool vis[55][55][1<<11];
struct Node {
int x,y;
int step;
int kk;
Node(int x,int y,int step,int kk):x(x),y(y),step(step),kk(kk){}
};
int bfs()
{
queue<Node> q;
memset(vis,0,sizeof(vis));
q.push(Node(1,1,0,key[1][1]));
vis[1][1][key[1][1] ] =1;
int nx,ny;
while(!q.empty()) {
Node cur = q.front();
q.pop();
if(cur.x == n && cur.y == m) return cur.step;
for(int k = 0; k<4; k++) {
nx = cur.x + tx[k];
ny = cur.y + ty[k];
int t = maze[cur.x][cur.y][nx][ny];
int nowkey = cur.kk | key[nx][ny];
int w = 1<<(t-1);
if(nx > n || ny > m || nx <1 || ny <1) continue;
if(vis[nx][ny][nowkey] == 1) continue;
if(t == 0 ) continue;
if(t!=-1 && (nowkey & w ) == 0) continue;
vis[nx][ny][nowkey] = 1;
q.push(Node(nx,ny,cur.step+1,nowkey));
}
}
return -1;
}
int main()
{
//n行m咧
int x1,x2,y1,y2;
int g,p,k;
while(~scanf("%d%d%d",&n,&m,&p) ) {
scanf("%d",&k);
memset(maze,-1,sizeof(maze));
memset(key,0,sizeof(key));
for(int i = 1; i<=k; i++) {
scanf("%d%d%d%d%d",&x1,&y1,&x2,&y2,&g);
maze[x1][y1][x2][y2] = maze[x2][y2][x1][y1] = g;
}
int s;
scanf("%d",&s);
while(s--) {
scanf("%d%d%d",&x1,&y1,&g);
g--;
key[x1][y1] |=(1<<g);
}
printf("%d\n",bfs());
}
return 0 ;
}
总结：
1.没事不要瞎定义变量，也不要瞎定义全局变量（因为有可能定义了然后错用了还不报错）！这题的bfs中我刚开始就定义了x，y，然后有一个地方应该是cur.x和cur.y我用成了x，y。。。本来只是落下了cur.但谁知编译器没报错！所以答案显然是错的。
2.if中该加括号的地方一定要加括号，你咋知道在评测机上的编译器上是怎么跑的程序。
3.类似g--这样的细节一定要注意