挑战程序2.1.5 穷竭搜索>>宽度优先搜索（练POJ3669）

 1 #include <stdio.h>
 2 #include <queue>
 3 #include <string.h>
 4 using namespace std;
 5 int map1[305][305],step[305][305];
 6 typedef pair<int, int> P;
 7 struct stu{
 8 int x; int y; int t;
 9 };
10 struct stu ch[50002];
11 
12 int dx[5]={0, 1, 0, -1, 0};
13 int dy[5]={0, 0, 1, 0, -1};
14 
15 void zha(int M) {
16     memset(map1, -1, sizeof(map1));
17     for(int i = 0; i < M; i++)
18         for(int k = 0; k < 5; k++){//five points are burst.
19             int nx = ch[i].x + dx[k];
20             int ny = ch[i].y + dy[k];
21             if(nx >= 0 && nx < 303 && ny >= 0 && ny < 303 && (ch[i].t < map1[ny][nx] || map1[ny][nx] == -1))
22                 map1[ny][nx] = ch[i].t;//give everypoint the minimum burst time.
23         }
24 }
25 
26 int bfs() {
27 queue <P> que;
28 que.push( P (0, 0));
29 memset(step, -1, sizeof(step));
30 step[0][0]=0;
31 while(que.size()) {
32     P p= que.front();
33     que.pop();
34         if(map1[p.first][p.second] == -1){//find map1‘s -1 and print it
35             return step[p.first][p.second];
36         }
37     if(step[p.first][p.second] >= map1[p.first][p.second])continue;//the point had benn burst,so jump it
38     for(int i = 1; i < 5; i++) {//this is the point of four diretions that does not been destroy.
39         int nx = p.first + dx[i],ny = p.second + dy[i];
40         if(nx>=0 && ny >=0 &&step[nx][ny] == -1)//if it can arrive and does not exceed the map
41             {
42                  que.push(P(nx, ny));//push it to the queue and go
43                  step[nx][ny]=step[p.first][p.second]+1;//of course,step must add 1 when go.
44             }
45     }
46 }
47 return -1;
48 }
49 
50 int main()
51 {
52     int M,a;
53     while(~scanf("%d",&M))
54     {
55          for(int i=0; i<M; i++)
56             scanf("%d%d%d",&ch[i].x,&ch[i].y,&ch[i].t);
57          zha(M);
58          a=bfs();
59          printf("%d\n",a);
60     }
61     return 0;
62 }