UVaLive 6585 && Gym 100299F Draughts (暴力+回溯)
amp 暴力 回溯 gym UVALive
2023-09-11 14:17:18 时间
题意:给定一个 10*10的矩阵,每一个W可以跳过一个B向对角走到#并把B吃掉,并且可以一直跳直到不能动为止,现在是W走的时候,问你最多吃几个B。
析:直接暴力+回溯,深搜就好。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 26 + 5; const int mod = 1e9 + 7; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } char s[15][15]; int ans; void dfs(int r, int c, int cnt){ ans = Max(ans, cnt); for(int i = 4; i < 8; ++i){ int x = r + dr[i]; int y = c + dc[i]; int xx = x + dr[i]; int yy = y + dc[i]; if(is_in(xx, yy) && s[x][y] == 'B' && s[xx][yy] == '#'){ s[x][y] = '#'; dfs(xx, yy, cnt+1); s[x][y] = 'B'; } } } int main(){ int T; cin >> T; n = m = 10; while(T--){ for(int i = 0; i < 10; ++i) scanf("%s", s+i); ans = 0; for(int i = 0; i < 10; ++i) for(int j = 0; j < 10; ++j) if(s[i][j] == 'W'){ s[i][j] = '#'; dfs(i, j, 0); s[i][j] = 'W'; } printf("%d\n", ans); } return 0; }
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