NYOJ 284 坦克大战 bfs + 优先队列

这类带权的边的图，直接广搜不行，要加上优先队列，这样得到的结果才是最优的，这样每次先找权值最小的，代码如下
  1 #include <stdio.h>
  2 #include <iostream>
  3 #include <queue>
  4 #include <string.h>
  5 using namespace std;
  6 typedef struct Node{
  7     int x, y;
  8     int step;
  9     friend bool operator < (const Node &a, const Node &b)
 10     {
 11         return a.step > b.step;
 12     }
 13 }Node;
 14 const int MAX = 300 + 5;
 15 int N, M;
 16 int Map[MAX][MAX];
 17 int next[4][2] = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
 18 Node s, e;
 19 bool match(Node node)//判断是否到达终点
 20 {
 21     if (node.x == e.x && node.y == e.y)
 22         return true;
 23     return false;
 24 }
 25 bool check(Node node)//判断这个点是否能走
 26 {
 27     if (node.x < 0 || node.y < 0 || node.x >= N || node.y >= M || Map[node.x][node.y] == 0)
 28         return false;
 29     return true;
 30 }
 31 int bfs()
 32 {
 33     priority_queue<Node> q;//优先队列
 34     q.push(s);
 35     Node p1, p2;
 36     while (!q.empty())
 37     {
 38         p1 = q.top();
 39         q.pop();
 40         for (int i = 0; i < 4; i++)
 41         {
 42             p2.x = p1.x + next[i][0];
 43             p2.y = p1.y + next[i][1];
 44             p2.step = p1.step + Map[p2.x][p2.y];
 45             if (match(p2))
 46             {
 47                 return p2.step;
 48             }
 49             if (check(p2))
 50             {
 51                 Map[p2.x][p2.y] = 0;
 52                 Node v = p2;
 53                 q.push(v);
 54             }
 55         }
 56     }
 57     return -1;
 58 }
 59 int main()
 60 {
 61     char ch;
 62     while (~scanf("%d%d", &N, &M) && N + M)
 63     {
 64         for (int i = 0; i < N; i++)
 65         {
 66             for (int j = 0; j < M; j++)
 67             {
 68                 cin >> ch;
 69                 if (ch == 'Y')//起点
 70                 {
 71                     s.x = i;
 72                     s.y = j;
 73                     s.step = 0;
 74                     Map[i][j] = 0;
 75                 }
 76                 else if (ch == 'T')//终点
 77                 {
 78                     e.x = i;
 79                     e.y = j;
 80                     e.step = 0;
 81                     Map[i][j] = 1;
 82                 }
 83                 else if (ch == 'B')//普通砖块，权值为2
 84                 {
 85                     Map[i][j] = 2;
 86                 }
 87                 else if (ch == 'E')//空地，权值为1
 88                 {
 89                     Map[i][j] = 1;
 90                 }
 91                 else
 92                 {
 93                     Map[i][j] = 0;//为0的时候表示此点不可走
 94                 }
 95             }
 96         }
 97         printf("%d
", bfs());
 98     }
 99 
100     return 0;
101 }