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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 }
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BZOJ1671 [Usaco2005 Dec]Knights of Ni 骑士