HDU 2722 Here We Go(relians) Again

就是题目长...  只要构好图就可以了。

1:each block is a square measuring 2520 rels per side 。

2:The speed limit between two adjacent intersections is always constant, and may range from 1 to 9 rels per blip

每条边的边长都是2520，速度1-9. 所以按time = 2520 / speed 构图。

 

00447988

2010-07-05 21:03:02

Accepted

1012

15 MS

1180 KB

Visual C++

Slave_wc

#include <iostream> #include <cstring> #include <queue> using namespace std; #define MAXN 500 const int slave = 2520; int map[MAXN][MAXN]; int dis[MAXN]; bool hash[MAXN]; struct Dij {     int id;     int dis;     bool friend operator < (const Dij a, const Dij b)     {         return a.dis > b.dis;     } }; Dij now, next; int Bfs(int st, int ed, int n) {     memset(hash, false, sizeof(hash));     memset(dis, 0x7f, sizeof(dis));     priority_queue<Dij> Q;     now.id = st;     now.dis = 0;     dis[st] = 0;     Q.push(now);          while (!Q.empty())     {         now = Q.top();         Q.pop();                  if (hash[now.id])         {             continue;         }         if (now.id == ed)         {             return now.dis;         }         hash[now.id] = true;                  for (int j = 1; j <= n; j++)         {             if (!hash[j] && dis[j] > dis[now.id] + map[now.id][j])             {                 dis[j] = dis[now.id] + map[now.id][j];                 next.id = j;                 next.dis = dis[j];                  Q.push(next);             }         }     }     return -1; } int main() {     int n, m, i, j, a, r, t;     char str[10];          while (scanf("%d %d", &n, &m) != EOF)     {         if (n == 0 && m == 0)         {             break;         }         r = 1;         memset(map, 0x7f, sizeof(map));         for (i = 1; i <= 2 * n + 1; i++)         {             if (i & 1)             {                 for (j = 1; j <= m; j++)                 {                     scanf("%d %s", &a, str);                     if (a == 0)                     {                         continue;                     }                     int left = (r - 1) * (m + 1) + j;                     int right = left + 1;                     if (str[0] == '*')                     {                         t = slave / a;                         map[left][right] = t;                         map[right][left] = t;                     }                     else if (str[0] == '>')                     {                         t = slave / a;                         map[left][right] = t;                          }                     else if (str[0] == '<')                     {                         t = slave / a;                         map[right][left] = t;                     }                 }                 r++;             }             else             {                 for (j = 1; j <= m + 1; j++)                 {                     scanf("%d %s", &a, str);                     if (a == 0)                     {                         continue;                     }                     int down = (r - 1) * (m + 1) + j;                     int up = down - (m + 1);                     if (str[0] == '*')                     {                         t = slave / a;                         map[up][down] = t;                         map[down][up] = t;                     }                     else if (str[0] == '^')                     {                         t = slave / a;                         map[down][up] = t;                          }                     else if (str[0] == 'v')                     {                         t = slave / a;                         map[up][down] = t;                     }                 }             }         }         int st = 1;         int ed = (n + 1) * (m + 1);         int ans = Bfs(st, ed, ed);         if (ans != -1)         {             printf("%d blips/n", ans);         }         else         {             puts("Holiday");         }     }     return 0; }