BZOJ3393 [Usaco2009 Jan]Laserphones 激光通讯

第一次知道。。原来spfa还可以这样写。。。用pq。。。

只需要直接求拐点即可，数据小想怎么搞就怎么搞

（话说怎么这么裸的最短路都写不出来了233）

 

 1 /**************************************************************
 2     Problem: 3393
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:36 ms
 7     Memory:1156 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <cstring>
12 #include <algorithm>
13 #include <queue>
14  
15 using namespace std;
16 const int dx[4] = {0, 1, 0, -1};
17 const int dy[4] = {1, 0, -1, 0};
18 const int N = 105;
19 const int inf = (int) 1e9;
20  
21 struct data {
22     int x, y, to, dis;
23     data() {}
24     data(int _x, int _y, int _t, int _d) : x(_x), y(_y), to(_t), dis(_d) {}
25      
26     inline bool operator < (const data &b) const {
27         return dis > b.dis;
28     }
29 };
30  
31 int n, m, ans;
32 int sx, sy, ex, ey;
33 int dis[N][N][4], w[N][N];
34 priority_queue <data> q;
35  
36 inline bool in(int x, int y) {
37     return 1 <= x && x <= n && 1 <= y && y <= m;
38 }
39  
40 int main() {
41     int i, j;
42     char ch;
43     scanf("%d%d", &m, &n);
44     for (i = 1; i <= n; ++i)
45         for (j = 1; j <= m; ++j) {
46         ch = getchar();
47         while (ch != 'C' && ch != '.' && ch != '*')
48             ch = getchar();
49         if (ch == '*') w[i][j] = 1;
50         if (ch == 'C')
51             if (!sx) sx = i, sy = j;
52             else ex = i, ey = j;
53     }
54     memset(dis, 127 / 3, sizeof(dis));
55     for (i = 0; i < 4; ++i) {
56         q.push(data(sx, sy, i, 0));
57         dis[sx][sy][i] = 0;
58     }
59      
60     int x, y, k, x1, y1;
61     while (!q.empty()) {
62         x1 = x = q.top().x, y1 = y = q.top().y, k = q.top().to;
63         while (in(x1 += dx[k], y1 += dy[k]) && !w[x1][y1] && dis[x1][y1][k] > dis[x][y][k]) {
64             dis[x1][y1][k] = dis[x][y][k];
65             q.push(data(x1, y1, k, dis[x][y][k]));
66         }
67         for (i = 0; i < 4; ++i)
68             if (dis[x][y][i] > q.top().dis + 1) {
69                 dis[x][y][i] = q.top().dis + 1;
70                 q.push(data(x, y, i, dis[x][y][i]));
71             }
72         q.pop();
73     }
74     for (i = 0, ans = inf; i < 4; ++i)
75         ans = min(ans, dis[ex][ey][i]);
76     printf("%d\n", ans);
77     return 0;
78 }
View Code

 

posted on 2014-12-02 17:18  Xs酱~  阅读(277)  评论(0)  编辑  收藏  举报