nyoj 坦克大战 (优先队列)

BFS题目。输入完图之后可以先预处理一下，然后用优先队列解决。

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstdlib>
 4 #include<cstring>
 5 #include<string>
 6 #include<queue>
 7 #include<algorithm>
 8 #include<map>
 9 #include<iomanip>
10 #include<climits>
11 #include<string.h>
12 #include<cmath>
13 #include<stdlib.h>
14 #include<vector>
15 #include<set>
16 #define INF 1e7
17 #define MAXN 100010
18 #define maxn 111
19 #define maxm 1000
20 #define Mod 1000007
21 #define MIN(a,b) (a < b ? a : b)
22 #define MAX(a,b) (a > b ? a : b)
23 #define mem(a) memset(a,0,sizeof(a))
24 using namespace std;
25 typedef long long LL;
26 int n, m;
27 int sx, sy, fx, fy;
28 char G[333][333];
29 int mp[333][333];
30 int vis[333][333];
31 int dx[] = { 1, -1, 0, 0 }, dy[] = { 0, 0, 1, -1 };
32
33 struct node {
34     int x, y, step;
35     bool operator <(const node a) const {
36         return step > a.step;
37     }
38 };
39
40 bool check(int x, int y)
41 {
42     if (vis[x][y] || x < 0 || y < 0 || x >= n || y >= m)
43         return true;
44     return false;
45 }
46
47 int bfs()
48 {
49     priority_queue<node> q;
50     q.push({ sx, sy, 0 });
51     vis[sx][sy] = 1;
52     while (!q.empty()) {
53         node now = q.top();
54         q.pop();
55         if (now.x == fx && now.y == fy)
56             return now.step;
57         node next;
58         for (int i = 0; i < 4; ++i) {
59             next.x = now.x + dx[i];
60             next.y = now.y + dy[i];
61             next.step = now.step;
62             if (check(next.x, next.y)) continue;
63             if (mp[next.x][next.y] != -1) {
64                 q.push({ next.x, next.y, next.step + mp[next.x][next.y] });
65                 vis[next.x][next.y] = 1;
66             }
67         }
68     }
69     return -1;
70 }
71
72 int main()
73 {
74     while (~scanf("%d%d",&n,&m), n + m) {
75         memset(vis,0,sizeof(vis));
76         memset(mp,0,sizeof(mp));
77         for (int i = 0; i < n; ++i) {
78             scanf("%s", G[i]);
79             for (int j = 0; j < m; ++j) {
80                 if (G[i][j] == ‘Y‘)
81                     sx = i, sy = j;
82                 else if (G[i][j] == ‘T‘)
83                     fx = i, fy = j;
84             }
85         }
86         for (int i = 0; i < n; ++i)
87             for (int j = 0; j < m; ++j) {
88                 if (G[i][j] == ‘S‘ || G[i][j] == ‘R‘) mp[i][j] = -1;
89                 else if (G[i][j] == ‘B‘) mp[i][j] = 2;
90                 else if (G[i][j] == ‘E‘ || G[i][j] == ‘T‘) mp[i][j] = 1;
91             }
92         cout << bfs() << endl;
93     }
94     return 0;
95 }