hihoCoder太阁最新面经算法竞赛19

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.

 def do(p1x, p1y, p2x, p2y, p3x, p3y):
     return (p3x - p1x) * (p2y - p1y) - (p2x - p1x) * (p3y - p1y);

 T = map(int, raw_input().split(' '))[0]
 while T > 0:
     T -= 1
     px, py, ax, ay, bx, by, cx, cy = map(int, raw_input().split(' '))
     x, y, z = do(ax, ay, px, py, bx, by), do(bx, by, px, py, cx, cy), do(cx, cy, px, py, ax, ay)
     if (x >= 0 and y >= 0 and z >= 0) or (x <= 0 and y <= 0 and z <= 0):
         print 'YES'
     else:
         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。

 #include <bits/stdc++.h>
 using namespace std;

 typedef long long LL;
 const LL mod = 1000000007LL;
 const int maxn = ;
 LL n;
 LL ret;

 typedef struct Matrix {
     LL m[maxn][maxn];
     int r;
     int c;
     Matrix() {
         r = c = ;
         memset(m, , sizeof(m));
     }
 } Matrix;

 Matrix mul(Matrix m1, Matrix m2, LL mod) {
     Matrix ans = Matrix();
     ans.r = m1.r;
     ans.c = m2.c;
     for(int i = ; i <= m1.r; i++) {
         for(int j = ; j <= m2.r; j++) {
                for(int k = ; k <= m2.c; k++) {
                 if(m2.m[j][k] == ) continue;
                 ans.m[i][k] = ((ans.m[i][k] + m1.m[i][j] * m2.m[j][k] % mod) % mod) % mod;
             }
         }
     }
     return ans;
 }

 Matrix quickmul(Matrix m, LL n, LL mod) {
     Matrix ans = Matrix();
     for(int i = ; i <= m.r; i++) {
         ans.m[i][i]  = ;
     }
     ans.r = m.r;
     ans.c = m.c;
     while(n) {
         if(n & ) {
             ans = mul(m, ans, mod);
         }
         m = mul(m, m, mod);
         n >>= ;
     }
     return ans;
 }

 void dfs(LL n, LL cur, LL sz) {
     if(n == ) {
         ret = max(ret, cur);
         return;
     }
     if(n >= ) dfs(n-,cur*%mod, cur);
     if(n >=  && sz) dfs(n-, (cur+sz)%mod, sz);
         dfs(n-,(cur+)%mod, sz);
 }

 int main() {
     // freopen("in", "r", stdin);
     while(cin >> n) {
         if(n < ) {
             ret = ;
             dfs(n, , );
             cout << ret << endl;
             continue;
         }
         Matrix x; x.r = , x.c = ;
         x.m[][] = , x.m[][] = , x.m[][] = , x.m[][] = , x.m[][] = ;
         Matrix p; p.r = p.c = ;
         memset(p.m, , sizeof(p.m));
         p.m[][] = ;
         for(int i = ; i <= ; i++) p.m[i][i-] = ;
         p = quickmul(p, n-, mod);
         p = mul(p, x, mod);
         cout << p.m[][] << endl;
     }
     return ;
 }

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。

 #include <bits/stdc++.h>
 using namespace std;

 const int maxn = ;
 const int maxm = ;
 const int maxc = ;

 typedef pair<int, int> pii;
 typedef struct Trie {
     int rt, sz;
     int id[maxm][maxc], val[maxm];
     int dp[maxm];
     string s;
     void build() {
         rt = sz = ;
         memset(dp, , sizeof(dp));
         memset(id, -, sizeof(id));
         memset(val, , sizeof(val));
     }
     void insert(const char* str) {
         int u = rt;
         int n = strlen(str);
         for(int i = ; i < n; i++) {
             int v = str[i] - 'a';
             if(id[u][v] == -) id[u][v] = ++sz;
             u = id[u][v];
         }
         val[u] = max(val[u], n);
     }
     void dfs(int x, int u) {
         if(x == s.length()) return;
         for(int i = ; i < ; i++) {
             int &v = id[u][i];
             if(v == -) continue;
             int y = x;
             while(s[y] != i + 'a' && y < s.length()) y++;
             if(y == s.length()) break;
             if(i + 'a' == s[y]) {
                 dp[v] = dp[u] + ;
                 dfs(y+, v);
             }
         }
     }
 }Trie;

 int n;
 char tmp[maxm];
 Trie trie;

 int main() {
     // freopen("in", "r", stdin);
     while(~scanf("%d", &n)) {
         trie.build();
         for(int i = ; i < n; i++) {
             scanf("%s", tmp);
             trie.insert(tmp);
         }
         scanf("%s", tmp);
         trie.s = tmp;
         trie.dfs(, );
         int ret = ;
         for(int i = ; i <= trie.sz; i++) {
             // printf("%d %d\n", trie.dp[i], trie.val[i]);
             if(trie.dp[i] != ) {
                 ret = max(ret, trie.val[i]);
             }
         }
         printf("%d\n", ret);
     }
     return ;
 }