题目传送门

解题思路:

二分答案,然后bfs验证,如果从一个路标可以达到其它所有路标,则答案可行.知道找到最佳答案.

AC代码:

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cmath>
 4 #include<cstring>
 5 #include<queue>
 6 
 7 using namespace std;
 8 
 9 int m,n,a[501][501],l,r,num,sum,ans,mid;
10 int dx[] = {-1,1,0,0},dy[] = {0,0,1,-1}; 
11 bool vis[501][501],v[501][501],flag,pp;
12 
13 inline bool check(int len,int x,int y) {
14     queue<int> q,q1;
15     q.push(x);
16     q1.push(y);
17     v[x][y] = 1;
18     while(!q.empty()) {
19         x = q.front();
20         y = q1.front();
21         q.pop();
22         q1.pop();
23         if(vis[x][y] == 1) num++;
24         if(num == sum) return true;
25         for(int i = 0;i < 4; i++) {
26             int xx = x + dx[i];
27             int yy = y + dy[i];
28             if(abs(a[xx][yy] - a[x][y]) > len) continue;
29             if(xx < 1 || xx > m || yy < 1 || yy > n) continue;
30             if(!v[xx][yy]) {
31                 q.push(xx);
32                 q1.push(yy);
33                 v[xx][yy] = 1;
34             }
35         }
36     }
37     return false;
38 }
39 
40 int main() {
41     int x,y;
42     scanf("%d%d",&m,&n);
43     for(int i = 1;i <= m; i++)
44         for(int j = 1;j <= n; j++){
45             scanf("%d",&a[i][j]);
46             r = max(r,a[i][j]);
47         }
48     for(int i = 1;i <= m; i++)
49         for(int j = 1;j <= n; j++) {
50             int u;
51             scanf("%d",&u);
52             if(u == 1) 
53                 vis[i][j] = 1,sum++,x = i,y = j;
54         }
55     while(l < r) {
56         mid = (l + r) / 2;
57         memset(v,0,sizeof(v));
58         flag = false;
59         num = 0;
60         if(check(mid,x,y))
61             r = mid,ans = mid;
62         else
63             l = mid + 1;
64     }
65     printf("%d",ans);
66     return 0;
67 }