比赛链接：http://hihocoder.com/contest/hihointerview28/problems
A. 固定一个方向，两两相邻的点顺时针或逆时针构造三个向量，判断这个点在这个向量的左侧还是右侧，看看是否在同一侧。trick就是点在向量上，对应的情况就是值为0.
1 def do(p1x, p1y, p2x, p2y, p3x, p3y): 2 return (p3x - p1x) * (p2y - p1y) - (p2x - p1x) * (p3y - p1y); 3 4 T = map(int, raw_input().split(‘ ‘))[0] 5 while T > 0: 6 T -= 1 7 px, py, ax, ay, bx, by, cx, cy = map(int, raw_input().split(‘ ‘)) 8 x, y, z = do(ax, ay, px, py, bx, by), do(bx, by, px, py, cx, cy), do(cx, cy, px, py, ax, ay) 9 if (x >= 0 and y >= 0 and z >= 0) or (x <= 0 and y <= 0 and z <= 0): 10 print ‘YES‘ 11 else: 12 print ‘NO‘
B.打表规律，发现<=16的时候可以暴搜，>16的时候f(n)=4*f(n-5)（？如果没记错的话），矩阵加速一下就行了。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 typedef long long LL;
5 const LL mod = 1000000007LL;
6 const int maxn = 10;
7 LL n;
8 LL ret;
9
10 typedef struct Matrix {
11 LL m[maxn][maxn];
12 int r;
13 int c;
14 Matrix() {
15 r = c = 0;
16 memset(m, 0, sizeof(m));
17 }
18 } Matrix;
19
20 Matrix mul(Matrix m1, Matrix m2, LL mod) {
21 Matrix ans = Matrix();
22 ans.r = m1.r;
23 ans.c = m2.c;
24 for(int i = 1; i <= m1.r; i++) {
25 for(int j = 1; j <= m2.r; j++) {
26 for(int k = 1; k <= m2.c; k++) {
27 if(m2.m[j][k] == 0) continue;
28 ans.m[i][k] = ((ans.m[i][k] + m1.m[i][j] * m2.m[j][k] % mod) % mod) % mod;
29 }
30 }
31 }
32 return ans;
33 }
34
35 Matrix quickmul(Matrix m, LL n, LL mod) {
36 Matrix ans = Matrix();
37 for(int i = 1; i <= m.r; i++) {
38 ans.m[i][i] = 1;
39 }
40 ans.r = m.r;
41 ans.c = m.c;
42 while(n) {
43 if(n & 1) {
44 ans = mul(m, ans, mod);
45 }
46 m = mul(m, m, mod);
47 n >>= 1;
48 }
49 return ans;
50 }
51
52 void dfs(LL n, LL cur, LL sz) {
53 if(n == 0) {
54 ret = max(ret, cur);
55 return;
56 }
57 if(n >= 3) dfs(n-3,cur*2%mod, cur);
58 if(n >= 1 && sz) dfs(n-1, (cur+sz)%mod, sz);
59 dfs(n-1,(cur+1)%mod, sz);
60 }
61
62 int main() {
63 // freopen("in", "r", stdin);
64 while(cin >> n) {
65 if(n < 16) {
66 ret = 0;
67 dfs(n, 0, 0);
68 cout << ret << endl;
69 continue;
70 }
71 Matrix x; x.r = 5, x.c = 1;
72 x.m[1][1] = 81, x.m[2][1] = 64, x.m[3][1] = 48, x.m[4][1] = 36, x.m[5][1] = 27;
73 Matrix p; p.r = p.c = 5;
74 memset(p.m, 0, sizeof(p.m));
75 p.m[1][5] = 4;
76 for(int i = 2; i <= 5; i++) p.m[i][i-1] = 1;
77 p = quickmul(p, n-15, mod);
78 p = mul(p, x, mod);
79 cout << p.m[1][1] << endl;
80 }
81 return 0;
82 }
C.对于trie上的每个节点u，求最小的整数x满足 节点u对应的字符串(trie上root->u的路径) 是 S[1..x]的子序列。
1 #include <bits/stdc++.h>
2 using namespace std;
3
4 const int maxn = 10100;
5 const int maxm = 100100;
6 const int maxc = 30;
7
8 typedef pair<int, int> pii;
9 typedef struct Trie {
10 int rt, sz;
11 int id[maxm][maxc], val[maxm];
12 int dp[maxm];
13 string s;
14 void build() {
15 rt = sz = 0;
16 memset(dp, 0, sizeof(dp));
17 memset(id, -1, sizeof(id));
18 memset(val, 0, sizeof(val));
19 }
20 void insert(const char* str) {
21 int u = rt;
22 int n = strlen(str);
23 for(int i = 0; i < n; i++) {
24 int v = str[i] - ‘a‘;
25 if(id[u][v] == -1) id[u][v] = ++sz;
26 u = id[u][v];
27 }
28 val[u] = max(val[u], n);
29 }
30 void dfs(int x, int u) {
31 if(x == s.length()) return;
32 for(int i = 0; i < 26; i++) {
33 int &v = id[u][i];
34 if(v == -1) continue;
35 int y = x;
36 while(s[y] != i + ‘a‘ && y < s.length()) y++;
37 if(y == s.length()) break;
38 if(i + ‘a‘ == s[y]) {
39 dp[v] = dp[u] + 1;
40 dfs(y+1, v);
41 }
42 }
43 }
44 }Trie;
45
46 int n;
47 char tmp[maxm];
48 Trie trie;
49
50 int main() {
51 // freopen("in", "r", stdin);
52 while(~scanf("%d", &n)) {
53 trie.build();
54 for(int i = 0; i < n; i++) {
55 scanf("%s", tmp);
56 trie.insert(tmp);
57 }
58 scanf("%s", tmp);
59 trie.s = tmp;
60 trie.dfs(0, 0);
61 int ret = 0;
62 for(int i = 0; i <= trie.sz; i++) {
63 // printf("%d %d\n", trie.dp[i], trie.val[i]);
64 if(trie.dp[i] != 0) {
65 ret = max(ret, trie.val[i]);
66 }
67 }
68 printf("%d\n", ret);
69 }
70 return 0;
71 }
时间： 2024-09-28 21:12:24