# HDU 4499 Cannon (brute force solution)

Source: Internet
Author: User

Test instructions: Given a n*m chess board, put some pieces, ask you can put a few guns (Chinese chess gun).

Analysis: Actually very simple, because the chessboard is 5*5 biggest, then the direct violence on the line, can be regarded as a line, very water, time is very short, only 62ms.

The code is as follows:

`#include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream > #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector > #include <map> #include <cctype>using namespace std; typedef long Long Ll;typedef pair<int, int> p;c onst int inf = 0x3f3f3f3f;const double inf = 0x3f3f3f3f3f3f3f;const double EPS = 1e-8;const int maxn = 1e4 + 5;const int D R[] = {0, 0,-1, 1};const int dc[] = {-1, 1, 0, 0};int N, m;inline bool is_in (int r, int c) {return R >= 0 &&amp ; R < n && C >= 0 && C < m;}    int a[30];int ans;bool isly (int x) {int b[] = {x+1, x-1, X+m, x-m};    int d[] = {x/m*m+m-1, x/m*m, n-1, 0};        int c[] = {1,-1, M,-m};        for (int i = 0; i < 4; ++i) {bool OK = false; for (int j = B[i]; (I & 1 J >= D[i]: J <= D[i]); J + = C[i]) {if (ok && a[j]! =-1) {if (a[j] = = 1) return false;            else break;        } else if (!ok && a[j]! =-1) ok = true; }} return true;    void Dfs (int x, int cnt) {ans = max (ans, CNT);        for (int i = x; i < n; ++i) {if (a[i]! =-1 | |!isly (i)) continue;        A[i] = 1;        DFS (i+1, cnt+1);    A[i] =-1;    }}int Main () {int C;        while (scanf ("%d%d%d", &n, &m, &c) = = 3) {n *= m;        Memset (A,-1, sizeof (a));            for (int i = 0; i < C; ++i) {int x, y;            scanf ("%d%d", &x, &y);        A[x*m+y] = 0;        } ans = 0;        DFS (0, 0);    printf ("%d\n", ans); } return 0;}`

HDU 4499 Cannon (brute force solution)

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