Codeforces Round #533 (Div. 2) D. Kilani and the Game(BFS)

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题目链接:https://codeforces.com/contest/1105/problem/D

题意:p 个人在 n * m 的地图上扩展自己的城堡范围,每次最多走 a_i 步(曼哈顿距离),按 1 ~ p 的顺序,问最后每个人占领的点的数量。

题解:用一个队列维护当前起点,用另一个队列模拟当前起点走 a_i 步可以到达的全部点。(60 ~ 63行关键代码,多源bfs,不可分开跑)

    数据:

    4 3 2
    2 1
    1..
    1..
    ..2
    ...

    output:10 2

 

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define ull unsigned long long
 5 #define mst(a,b) memset((a),(b),sizeof(a))
 6 #define mp(a,b) make_pair(a,b)
 7 #define pi acos(-1)
 8 #define pii pair<int,int>
 9 #define pb push_back
10 #define lowbit(x) ((x)&(-x))
11 const int INF = 0x3f3f3f3f;
12 const double eps = 1e-6;
13 const int maxn = 1e3 + 10;
14 const int maxm = 1e6 + 10;
15 const ll mod =  1e9 + 7;
16 
17 int n,m,p;
18 char s[maxn][maxn];
19 int a[15],vis[maxn][maxn];
20 int dx[4] = {-1,0,0,1};
21 int dy[4] = {0,-1,1,0};
22 
23 struct node {
24     int x,y,d;
25 };
26 
27 bool judge(int x,int y) {
28     if(x <= 0 || x > n || y <= 0 || y > m || s[x][y] == #) return false;
29     return true;
30 }
31 
32 int main() {
33 #ifdef local
34     freopen("data.txt", "r", stdin);
35 //    freopen("data.txt", "w", stdout);
36 #endif
37     scanf("%d%d%d",&n,&m,&p);
38     for(int i = 1; i <= p; i++) scanf("%d",&a[i]);
39     for(int i = 1; i <= n; i++) {
40         scanf("%s",s[i] + 1);
41         for(int j = 1; j <= m; j++) vis[i][j] = -1;
42     }
43     queue<pii>q;
44     for(int k = 1; k <= p; k++) {
45         for(int i = 1; i <= n; i++) {
46             for(int j = 1; j <= m; j++) {
47                 if(s[i][j] - 0 == k) {
48                     q.push(mp(i,j));
49                     vis[i][j] = k;
50                 }
51             }
52         }
53     }
54     while(!q.empty()) {
55         pii now = q.front();
56         q.pop();
57         int x = now.first, y = now.second;
58         queue<node>q1;
59         q1.push({x,y,a[vis[x][y]]});
60         while(!q.empty() && vis[q.front().first][q.front().second] == vis[x][y]) {
61             q1.push({q.front().first,q.front().second,a[vis[x][y]]});
62             q.pop();
63         }
64         while(!q1.empty()) {
65             node hh = q1.front();
66             q1.pop();
67             if(hh.d <= 0) continue;
68             for(int i = 0; i < 4; i++) {
69                 int nx = hh.x + dx[i], ny = hh.y + dy[i], d = hh.d;
70                 if(judge(nx,ny) && vis[nx][ny] == -1) {
71                     vis[nx][ny] = vis[x][y];
72                     q1.push({nx,ny,d - 1});
73                     q.push(mp(nx,ny));
74                 }
75             }
76         }
77     }
78     int ans[15] = {0};
79     for(int i = 1; i <= n; i++) {
80         for(int j = 1; j <= m; j++) {
81             if(vis[i][j] != -1) ans[vis[i][j]]++;
82         }
83     }
84     for(int i = 1; i <= p; i++) printf("%d ",ans[i]);
85     return 0;
86 }

 






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