BZOJ1671 [Usaco2005 Dec]Knights of Ni 骑士

打水题啊打水题。。。然后就被虐了。。。

看不懂题目，后来查hzwer的blog，发现时题目翻译错了。。。我去我就说我的语文怎么这么差。

hzwer："总之就是在地图上从2出发，走到3，途中要至少经过一个4。"

 

原来如此，不就是两次bfs嘛。。。只是第一次bfs的时候要注意不能经过4。

然后开始bfs，在快要写死的时候。。。终于对了。。。

哈哈，status第六哦~~~不错不错

 

 1 /**************************************************************
 2     Problem: 1671
 3     User: rausen
 4     Language: C++
 5     Result: Accepted
 6     Time:168 ms
 7     Memory:16548 kb
 8 ****************************************************************/
 9  
10 #include <cstdio>
11 #include <algorithm>
12  
13 using namespace std;
14 const int inf = (int) 1e9;
15 const int dx[4] = {0, 0, -1, 1};
16 const int dy[4] = {1, -1, 0, 0};
17  
18 struct QUEUE{
19     int x, y;
20 }q[500005];
21 int d[2][1005][1005], mp[1005][1005];
22 int n, m, sx[2], sy[2];
23  
24 inline int read(){
25     int x = 0, sgn = 1;
26     char ch = getchar();
27     while (ch < '0' || ch > '9'){
28         if (ch == '-') sgn = -1;
29         ch = getchar();
30     }
31     while (ch >= '0' && ch <= '9'){
32         x = x * 10 + ch - '0';
33         ch = getchar();
34     }
35     return sgn * x;
36 }
37  
38 inline bool check(int x, int y){
39     return x < 1 || y < 1 || x > n || y > m;
40 }
41  
42 void bfs(int p){
43     q[0].x = sx[p], q[0].y = sy[p], d[p][sx[p]][sy[p]] = 0;
44     int x, y, X, Y;
45     for (int l = 0, r = 0; l <= r; ++l){
46         X = q[l].x, Y = q[l].y;
47         for (int i = 0; i < 4; ++i){
48             x = X + dx[i], y = Y + dy[i];
49             if (check(x, y) || mp[x][y] == 1 || d[p][x][y] != -1) continue;
50             ++r;
51             q[r].x = x, q[r].y = y;
52             d[p][x][y] = d[p][X][Y] + 1;
53         }
54     }
55 }
56  
57 int getans(){
58     int res = inf;
59     for (int i = 1; i <= n; ++i)
60         for (int j = 1; j <= m; ++ j){
61             if (d[0][i][j] == -1 || d[1][i][j] == -1 || mp[i][j] != 4) continue;
62             res = min(res, d[0][i][j] + d[1][i][j]);
63         }
64     return res;
65 }
66  
67 int main(){
68     m = read(), n = read();
69     for (int i = 1; i <= n; ++i)
70         for (int j = 1; j <= m; ++j){
71             d[0][i][j] = d[1][i][j] = -1;
72             mp[i][j] = read();
73             if (mp[i][j] == 2)
74                 sx[0] = i, sy[0] = j;
75             if (mp[i][j] == 3)
76                 sx[1] = i, sy[1] = j;
77         }
78          
79     bfs(0); bfs(1);
80     printf("%d\n", getans());
81     return 0;
82 }
View Code

 

posted on 2014-10-19 20:43  Xs酱~  阅读(288)  评论(0)  编辑  收藏  举报