链接:https://codeforces.com/contest/1105/problem/D
题意:
给n*m的地图,最多9个人,同时有每个人的扩张次数(我开始以为是直线扩张最大长度。。实际是能连续扩张次数。)
地图上有‘#’,‘.',和数字,数字对应每个人的据点,
从1-n轮流扩张。
地图被扩张完后,输入每个人的据点数目。
思路:
赛后写的题。还一堆bug,
用队列和一个数组,记录每个人能扩张的点和下一次能扩张的个数。
然后就是一堆循环套着。每次入队更新下一次的扩张个数,同时用flag记录有几个人还可以扩张。
不能扩张就减一。
代码:
#includeusing namespace std;const int MAXN = 1010;struct Node{ int _x; int _y; Node(int x,int y):_x(x),_y(y){}};int Next[4][2] = { {-1,0},{0,1},{1,0},{0,-1}};int next_time[10];int Map[MAXN][MAXN];int number[10];int speed[10];int vis[10];queue player[10];int main(){ int n, m, p; char c; scanf("%d%d%d", &n, &m, &p); for (int i = 1; i <= p; i++) { scanf("%d", &speed[i]); } for (int i = 1; i <= n; i++) { for (int j = 1; j <= m; j++) { cin >> c; if (c == '.') Map[i][j] = 0; else if (c == '#') Map[i][j] = -1; else { Map[i][j] = c - '0'; player[c - '0'].push(Node(i, j)); number[c - '0']++; next_time[c - '0']++; } } } int flag = p; while (flag > 0) { for (int i = 1; i <= p; i++) { if (player[i].empty()) continue; for (int times = 1; times <= speed[i]; times++) { int ti = next_time[i]; next_time[i] = 0; for (int z = 1; z <= ti; z++) { //cout << 2 << endl; int x = player[i].front()._x; int y = player[i].front()._y; player[i].pop(); for (int j = 0; j < 4; j++) { int tx = x + Next[j][0]; int ty = y + Next[j][1]; if (tx < 1 || tx > n || ty < 1 || ty > m) continue; if (Map[tx][ty] != 0 || Map[tx][ty] == '#') continue; Map[tx][ty] = i; number[i]++; player[i].push(Node(tx, ty)); next_time[i]++; } } if (player[i].empty()&&vis[i] == 0) { flag--; vis[i] = 1; break; } /* for (int v = 1;v<=n;v++) { for (int c = 1;c<=m;c++) cout << Map[v][c] << ' '; cout << endl; } */ } //cout << 3 << endl; //cout << player[i].size() << endl; } if (flag <= 0) break; //cout << 4 << endl; } for (int i = 1; i <= p; i++) printf("%d ", number[i]); printf("\n"); return 0;}