hdu 4414 Finding Crosses && hdu 4417 Super Mario
杭州网络赛的两道水题，4414是暴力搜索，找到满足条件的十字架，4417是简单的线段树，需要查找区间中满足条件的数的个数。
代码如下：
hdu 4414
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 5 using namespace std; 6 7 const int maxn = 55; 8 int mp[2][maxn][maxn]; 9 char rec[maxn][maxn]; 10 11 void cntVer(int n) { 12 memset(mp[0], -1, sizeof(mp[0])); 13 14 int cnt = 0; 15 16 for (int i = 0; i < n; i++) { 17 cnt = 0; 18 for (int j = 0; j < n; j++) { 19 if (rec[i][j] == 'o') { 20 if ((cnt & 1) && cnt != 1) { 21 mp[0][i][j - (cnt >> 1) - 1] = cnt >> 1; 22 } 23 cnt = 0; 24 } else { 25 cnt++; 26 } 27 } 28 if ((cnt & 1) && cnt != 1) mp[0][i][n - (cnt >> 1) - 1] = cnt >> 1; 29 } 30 31 // puts("ver"); 32 // for (int i = 0; i < n; i++) { 33 // for (int j = 0; j < n; j++) { 34 // printf("%d ", mp[0][i][j]); 35 // } 36 // puts(""); 37 // } 38 } 39 40 void cntHor(int n) { 41 memset(mp[1], -1, sizeof(mp[1])); 42 43 int cnt = 0; 44 45 for (int i = 0; i < n; i++) { 46 cnt = 0; 47 for (int j = 0; j < n; j++) { 48 if (rec[j][i] == 'o') { 49 if ((cnt & 1) && cnt != 1) { 50 mp[1][j - (cnt >> 1) - 1][i] = cnt >> 1; 51 } 52 cnt = 0; 53 } else { 54 cnt++; 55 } 56 } 57 if ((cnt & 1) && cnt != 1) mp[1][n - (cnt >> 1) - 1][i] = cnt >> 1; 58 } 59 60 // puts("hor"); 61 // for (int i = 0; i < n; i++) { 62 // for (int j = 0; j < n; j++) { 63 // printf("%d ", mp[1][i][j]); 64 // } 65 // puts(""); 66 // } 67 } 68 69 bool judge(int x, int y) { 70 if (mp[0][x][y] != mp[1][x][y]) return false; 71 for (int i = 1; i <= mp[0][x][y]; i++) { 72 if (rec[x + 1][y + i] == '#' || rec[x - 1][y + i] == '#') return false; 73 if (rec[x + 1][y - i] == '#' || rec[x - 1][y - i] == '#') return false; 74 } 75 for (int i = 1; i <= mp[1][x][y]; i++) { 76 if (rec[x + i][y + 1] == '#' || rec[x + i][y - 1] == '#') return false; 77 if (rec[x - i][y - 1] == '#' || rec[x - i][y + 1] == '#') return false; 78 } 79 80 return true; 81 } 82 83 int deal(int n) { 84 int cnt = 0; 85 86 for (int i = 1; i < n - 1; i++) { 87 for (int j = 1; j < n - 1; j++) { 88 if (~mp[0][i][j] && ~mp[1][i][j] && judge(i, j)) cnt++; 89 } 90 } 91 92 return cnt; 93 } 94 95 int main() { 96 int n; 97 98 // freopen("in", "r", stdin); 99 while (~scanf("%d", &n) && n) { 100 for (int i = 0; i < n; i++) { 101 scanf("%s", rec[i]); 102 } 103 cntVer(n); 104 cntHor(n); 105 printf("%d\n", deal(n)); 106 } 107 108 return 0; 109 }
hdu 4417
1 #include <cstdio> 2 #include <vector> 3 #include <algorithm> 4 #include <cstring> 5 6 #define lson l, m, rt << 1 7 #define rson m + 1, r, rt << 1 | 1 8 9 using namespace std; 10 typedef vector<int> vi; 11 typedef pair<vi::iterator, vi::iterator> pii; 12 const int maxn = 100001; 13 14 vi rec[maxn << 2]; 15 int ht[maxn]; 16 17 void build(int l, int r, int rt){ 18 rec[rt].clear(); 19 for (int i = l; i <= r; i++){ 20 rec[rt].push_back(ht[i]); 21 } 22 int m = (l + r) >> 1; 23 24 sort(rec[rt].begin(), rec[rt].end()); 25 // for (vi::iterator ii = rec[rt].begin(); ii != rec[rt].end(); ii++){ 26 // printf("%d ", *ii); 27 // } 28 // puts("end\n"); 29 if (l == r) return ; 30 31 build(lson); 32 build(rson); 33 } 34 35 int query(int L, int R, int key, int l, int r, int rt){ 36 if (L <= l && r <= R){ 37 pii tmp; 38 39 tmp = equal_range(rec[rt].begin(), rec[rt].end(), key); 40 return tmp.second - rec[rt].begin(); 41 } 42 int m = (l + r) >> 1; 43 int ret = 0; 44 45 if (L <= m) ret = query(L, R, key, lson); 46 if (m < R) ret += query(L, R, key, rson); 47 48 return ret; 49 } 50 51 int main(){ 52 int n, m; 53 int T; 54 55 scanf("%d", &T); 56 for (int i = 1; i <= T; i++){ 57 scanf("%d%d", &n, &m); 58 for (int j = 0; j < n; j++){ 59 scanf("%d", &ht[j]); 60 } 61 printf("Case %d:\n", i); 62 build(0, n - 1, 1); 63 for (int j = 0; j < m; j++){ 64 int l, r, k; 65 66 scanf("%d%d%d", &l, &r, &k); 67 printf("%d\n", query(l, r, k, 0, n - 1, 1)); 68 } 69 } 70 71 return 0; 72 }
——written by Lyon
