Replay:
Dup4:
- 时间复杂度算不对？ 一点点思路不经过验证就激动的要死？ 浪费自己一个小时还浪费别人一个小时？
- 对1e3不敏感？ 1e3 * 1e3是多少？ 模拟建边跑dp不写非要写个大模拟？
- 看到数据结构就高兴的要死？ 没细想？ 没发现性质？
X:
- 日常语文差， 导致计算几何死都写不对 读题要细致啊！
- 感觉状态还可以?只是计算几何写太久了， 人都懵了
A：Cactus Draw
Solved.
按照BFS序以及深度排
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 const int maxn = 1e4 + 10; 6 7 struct Edge{ 8 int to, nxt; 9 Edge(){} 10 Edge(int to, int nxt) :to(to), nxt(nxt){} 11 }edge[maxn << 1]; 12 13 struct node{ 14 int x, y; 15 node(){} 16 node(int x, int y):x(x), y(y){} 17 }ans[maxn]; 18 19 int n, m; 20 int head[maxn], tot; 21 int vis[maxn]; 22 int level[maxn]; 23 24 void Init() 25 { 26 tot = 0; 27 memset(vis, 0, sizeof vis); 28 memset(level, 0, sizeof level); 29 memset(head, -1, sizeof head); 30 } 31 32 void addedge(int u,int v) 33 { 34 edge[tot] = Edge(v, head[u]); head[u] = tot++; 35 edge[tot] = Edge(u, head[v]); head[v] = tot++; 36 } 37 38 void BFS(int root) 39 { 40 queue<int>q; 41 q.push(root); 42 vis[root] = 1; 43 ans[root] = node(vis[root], ++level[vis[root]]); 44 while(!q.empty()) 45 { 46 int u = q.front(); 47 q.pop(); 48 for(int i = head[u]; ~i; i = edge[i].nxt) 49 { 50 int v = edge[i].to; 51 if(!vis[v]) 52 { 53 vis[v] = vis[u] + 1; 54 ans[v] = node(vis[v], ++level[vis[v]]); 55 q.push(v); 56 } 57 } 58 } 59 } 60 61 int main() 62 { 63 while(~scanf("%d %d", &n, &m)) 64 { 65 Init(); 66 for(int i = 1, u, v; i <= m; ++i) 67 { 68 scanf("%d %d", &u, &v); 69 addedge(u, v); 70 } 71 BFS(1); 72 for(int i= 1 ; i <= n; ++i) 73 { 74 printf("%d %d ", ans[i].x, ans[i].y); 75 } 76 } 77 return 0; 78 }
C：Division
Solved.
每次取最大进行操作，堆维护
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 100010 6 int n, k; 7 8 int main() 9 { 10 while (scanf("%d%d", &n, &k) != EOF) 11 { 12 priority_queue <int> pq; 13 ll res = 0; 14 for (int i = 1, a; i <= n; ++i) 15 { 16 scanf("%d", &a); 17 pq.push(a); 18 } 19 for (int i = 1; i <= k; ++i) 20 { 21 int top = pq.top(); pq.pop(); 22 pq.push(top / 2); 23 } 24 while (!pq.empty()) 25 { 26 res += pq.top(); 27 pq.pop(); 28 } 29 printf("%lld ", res); 30 } 31 return 0; 32 }
D：doppelblock
unsolved.
搜索。
增加剪枝：当剩余的数字小于x之间的数字时，回溯掉即可。
还有一条剪枝可以先处理x的位置在填数字（未写）
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const ll MOD = 1e9 + 7; 8 9 const int maxn = 1e2 + 10; 10 11 int n; 12 int sum = 0; 13 int r[maxn], c[maxn]; 14 int left_r[maxn], left_c[maxn]; 15 int mid_r[maxn], mid_c[maxn]; 16 int right_r[maxn], right_c[maxn]; 17 int cnt_r[maxn], cnt_c[maxn]; 18 int vis_r[maxn][maxn], vis_c[maxn][maxn]; 19 char mp[maxn][maxn]; 20 21 void Init() 22 { 23 memset(left_r, 0, sizeof left_r); 24 memset(left_c, 0, sizeof left_c); 25 26 memset(mid_r, 0, sizeof mid_r); 27 memset(mid_c, 0, sizeof mid_c); 28 29 memset(right_r, 0, sizeof right_r); 30 memset(right_c, 0, sizeof right_c); 31 32 memset(cnt_r, 0, sizeof cnt_r); 33 memset(cnt_c, 0, sizeof cnt_c); 34 35 memset(vis_r, 0, sizeof vis_r); 36 memset(vis_c, 0, sizeof vis_c); 37 } 38 39 bool DFS(int x, int y) 40 { 41 if (x == n && y == n + 1) return true; 42 if (y == n + 1) 43 { 44 if (cnt_r[x] != 2) return false; 45 else return DFS(x + 1, 1); 46 } 47 48 if (cnt_r[x] == 0 && cnt_c[y] == 0) 49 { 50 mp[x][y] = 'X'; 51 cnt_r[x]++; 52 cnt_c[y]++; 53 if (DFS(x, y + 1)) return true; 54 cnt_r[x]--; 55 cnt_c[y]--; 56 } 57 58 if (cnt_r[x] == 0 && (cnt_c[y] == 1 && mid_c[y] == c[y])) 59 { 60 mp[x][y] = 'X'; 61 cnt_r[x]++; 62 cnt_c[y]++; 63 if (DFS(x, y + 1)) return true; 64 cnt_r[x]--; 65 cnt_c[y]--; 66 } 67 68 if (cnt_c[y] == 0 && (cnt_r[x] == 1 && mid_r[x] == r[x])) 69 { 70 mp[x][y] = 'X'; 71 cnt_r[x]++; 72 cnt_c[y]++; 73 if (DFS(x, y + 1)) return true; 74 cnt_r[x]--; 75 cnt_c[y]--; 76 } 77 78 if ((cnt_r[x] == 1 && mid_r[x] == r[x]) && (cnt_c[y] == 1 && mid_c[y] == c[y])) 79 { 80 mp[x][y] = 'X'; 81 cnt_r[x]++; 82 cnt_c[y]++; 83 if (DFS(x, y + 1)) return true; 84 cnt_r[x]--; 85 cnt_c[y]--; 86 } 87 88 89 90 for (int i = 1; i <= n - 2; ++i) 91 { 92 if (vis_r[x][i] || vis_c[y][i]) continue; 93 if (cnt_r[x] == 0) 94 { 95 if (sum - (left_r[x] + i) < r[x]) continue; 96 } 97 else if (cnt_r[x] == 1) 98 { 99 if (mid_r[x] + i > r[x]) continue; 100 } 101 102 if (cnt_c[y] == 0) 103 { 104 if (sum - (left_c[y] + i) < c[y]) continue; 105 } 106 else if (cnt_c[y] == 1) 107 { 108 if (mid_c[y] + i > c[y]) continue; 109 } 110 111 if (cnt_r[x] == 0) left_r[x] += i; 112 else if (cnt_r[x] == 1) mid_r[x] += i; 113 else if (cnt_r[x] == 2) right_r[x] += i; 114 115 if (cnt_c[y] == 0) left_c[y] += i; 116 else if (cnt_c[y] == 1) mid_c[y] += i; 117 else if (cnt_c[y] == 2) right_c[y] += i; 118 119 vis_r[x][i]++; 120 vis_c[y][i]++; 121 122 mp[x][y] = i + '0'; 123 if (DFS(x, y + 1)) return true; 124 125 if (cnt_r[x] == 0) left_r[x] -= i; 126 else if (cnt_r[x] == 1) mid_r[x] -= i; 127 else if (cnt_r[x] == 2) right_r[x] -= i; 128 129 if (cnt_c[y] == 0) left_c[y] -= i; 130 else if (cnt_c[y] == 1) mid_c[y] -= i; 131 else if (cnt_c[y] == 2) right_c[y] -= i; 132 133 vis_r[x][i]--; 134 vis_c[y][i]--; 135 } 136 137 return false; 138 } 139 140 void RUN() 141 { 142 int t; 143 int flag = 0; 144 scanf("%d", &t); 145 while (t--) 146 { 147 if (flag++) printf(" "); 148 Init(); 149 scanf("%d", &n); 150 for (int i = 1; i <= n; ++i) scanf("%d", r + i); 151 for (int i = 1; i <= n; ++i) scanf("%d", c + i); 152 sum = (n - 2) * (n - 1) / 2; 153 DFS(1, 1); 154 for (int i = 1; i <= n; ++i) 155 { 156 for (int j = 1; j <= n; ++j) 157 { 158 printf("%c", mp[i][j]); 159 } 160 puts(""); 161 } 162 } 163 } 164 165 int main() 166 { 167 #ifdef LOCAL_JUDGE 168 freopen("Text.txt", "r", stdin); 169 #endif // LOCAL_JUDGE 170 171 RUN(); 172 173 #ifdef LOCAL_JUDGE 174 fclose(stdin); 175 #endif // LOCAL_JUDGE 176 return 0; 177 }
E：Fast Kronecker Transform
Upsolved.
将同样的数放在一起，如果同样的数字小于$10000，直接暴力$
否则做NTT
$因为模数是998244353，可以直接做，做FFT可能有精度问题$
$F(n) = sum f(t) cdot g(n - t)$
$f(t) = t 当 a_t = x$
$g(t) = t 当 b_t = x$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define db long double 5 #define ll long long 6 #define N 400010 7 #define S 10010 8 const ll MOD = (ll)998244353; 9 int n, m, a[N], b[N], c[N]; 10 vector <int> l[N], r[N]; 11 ll ans[N]; 12 13 void Hash() 14 { 15 c[0] = 0; 16 for (int i = 0; i <= n; ++i) c[++c[0]] = a[i]; 17 for (int i = 0; i <= m; ++i) c[++c[0]] = b[i]; 18 sort(c + 1, c + 1 + c[0]); 19 c[0] = unique(c + 1, c + 1 + c[0]) - c - 1; 20 for (int i = 0; i <= n; ++i) a[i] = lower_bound(c + 1, c + 1 + c[0], a[i]) - c; 21 for (int i = 0; i <= m; ++i) b[i] = lower_bound(c + 1, c + 1 + c[0], b[i]) - c; 22 } 23 24 ll qmod(ll base, ll n) 25 { 26 ll res = 1; 27 while (n) 28 { 29 if (n & 1) res = (res * base) % MOD; 30 base = base * base % MOD; 31 n >>= 1; 32 } 33 return res; 34 } 35 36 int x1[N], x2[N]; 37 void ntt(int *a, int len, int f) 38 { 39 int i, j = 0, t, k; 40 for (int i = 1; i < len - 1; ++i) 41 { 42 for (t = len; j ^= t >>= 1, ~j & t;); 43 if (i < j) swap(a[i], a[j]); 44 } 45 for (int i = 1; i < len; i <<= 1) 46 { 47 t = i << 1; 48 int wn = qmod(3, (MOD - 1) / t); 49 for (int j = 0; j < len; j += t) 50 { 51 int w = 1; 52 for (k = 0; k < i; ++k, w = 1ll * w * wn % MOD) 53 { 54 int x = a[j + k], y = 1ll * w * a[j + k + i] % MOD; 55 a[j + k] = (x + y) % MOD, a[j + k + i] = (x - y + MOD) % MOD; 56 } 57 } 58 } 59 if (f == -1) 60 { 61 reverse(a + 1, a + len); 62 int inv = qmod(len, MOD - 2); 63 for (int i = 0; i < len; ++i) a[i] = 1ll * a[i] * inv % MOD; 64 } 65 } 66 67 int main() 68 { 69 while (scanf("%d%d", &n, &m) != EOF) 70 { 71 for (int i = 0; i <= n; ++i) scanf("%d", a + i); 72 for (int i = 0; i <= m; ++i) scanf("%d", b + i); Hash(); 73 for (int i = 0; i <= n; ++i) l[a[i]].push_back(i); 74 for (int i = 0; i <= m; ++i) r[b[i]].push_back(i); 75 int len1 = n + 1, len2 = m + 1, len = 1; 76 while (len < (len1 + len2)) len <<= 1; 77 memset(ans, 0, sizeof ans); 78 for (int i = 1; i <= n + m + 5; ++i) 79 { 80 if (l[i].size() + r[i].size() < S) 81 { 82 for (auto u : l[i]) for (auto v : r[i]) 83 ans[u + v] = (ans[u + v] + (1ll * u * v) % MOD) % MOD; 84 } 85 else 86 { 87 for (int j = 0; j < len; ++j) x1[j] = 0; 88 for (int j = 0; j < len; ++j) x2[j] = 0; 89 for (auto x : l[i]) x1[x] = x; 90 for (auto x : r[i]) x2[x] = x; 91 ntt(x1, len, 1); 92 ntt(x2, len, 1); 93 for (int j = 0; j < len; ++j) 94 x1[j] = 1ll * x1[j] * x2[j] % MOD; 95 ntt(x1, len, -1); 96 for (int j = 0; j <= n + m; ++j) 97 ans[j] = (ans[j] + x1[j]) % MOD; 98 } 99 } 100 for (int i = 0; i <= n + m; ++i) printf("%lld%c", ans[i] % MOD, " "[i == n + m]); 101 } 102 return 0; 103 }
F：Kropki
Solved.
习惯性记忆化搜索（实际上是个状压dp）
$dp[S][i]表示S状态下i作为最后一个出现的状态， dp下去即可$
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const ll MOD = 1e9 + 7; 8 9 int n; 10 ll dp[1 << 16][20]; 11 char str[20]; 12 13 ll DFS(int S, int last, int dep) 14 { 15 if(dep == n) 16 { 17 return 1ll; 18 } 19 if(dp[S][last] != -1) return dp[S][last]; 20 ll res = 0; 21 for(int i = 1; i <= n; ++i) 22 { 23 if(S & (1 << (i - 1))) continue; 24 if(dep) 25 { 26 if(str[dep] == '1') 27 { 28 if(i != last * 2 && i * 2 != last) continue; 29 } 30 if(str[dep] == '0') 31 { 32 if(i == last * 2 || i * 2 == last) continue; 33 } 34 } 35 ll tmp = DFS((S | (1 << (i - 1))), i, dep + 1); 36 res = (res + tmp) % MOD; 37 } 38 dp[S][last] = res; 39 return res; 40 } 41 42 int main() 43 { 44 while(~scanf("%d", &n)) 45 { 46 scanf("%s", str + 1); 47 memset(dp, -1, sizeof dp); 48 ll ans = DFS(0, 0, 0); 49 printf("%lld ", ans); 50 } 51 return 0; 52 }
H：Nested Tree
Solved.
点数只有$10^6，建边树形DP$
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 typedef long long ll; 6 7 const ll MOD = (ll)1e9 + 7; 8 const int maxn = 1e6 + 10; 9 10 struct Edge{ 11 int to, nxt; 12 Edge(){} 13 Edge(int to, int nxt):to(to), nxt(nxt){} 14 }edge[maxn << 1]; 15 16 int n, m; 17 int head[maxn], tot; 18 ll son[maxn]; 19 ll ans; 20 21 22 void Init() 23 { 24 ans = 0; 25 tot = 0; 26 memset(head, -1, sizeof head); 27 } 28 29 void addedge(int u,int v) 30 { 31 edge[tot] = Edge(v, head[u]); head[u] = tot++; 32 edge[tot] = Edge(u, head[v]); head[v] = tot++; 33 } 34 35 void DFS(int u, int fa) 36 { 37 son[u] = 1; 38 for(int i = head[u]; ~i; i = edge[i].nxt) 39 { 40 int v = edge[i].to; 41 if(v == fa) continue; 42 DFS(v, u); 43 son[u] += son[v]; 44 ans = (ans + (son[v] * (n - son[v]) % MOD) % MOD) % MOD; 45 } 46 } 47 48 int main() 49 { 50 while(~scanf("%d %d", &n, &m)) 51 { 52 Init(); 53 for(int i = 1, u, v; i < n; ++i) 54 { 55 scanf("%d %d", &u, &v); 56 for(int j = 1; j <= m; ++j) 57 { 58 addedge((j - 1) * n + u, (j - 1) * n + v); 59 } 60 } 61 for(int i = 1, a, b, u, v; i < m; ++i) 62 { 63 scanf("%d %d %d %d", &a ,&b, &u, &v); 64 addedge((a - 1) * n + u, (b - 1) * n + v); 65 } 66 n *= m; 67 DFS(1, -1); 68 printf("%lld ", ans); 69 } 70 return 0; 71 }
I：Sorting
Upsolved.
将数分为两类，一类是$<= x， 二类是> x $
同一类的数在怎么操作其相对位置都是不变的
那么我们只需要知道前缀区间内有多少个一类数，有多少个二类数
再用前缀和维护同一类数的和即可
$2、3操作用线段树维护即可，用0, 1分别表示一类数$
每次操作相当于将前面连续一段赋值为$0/1 后面连续一段赋值为1/0$
1 #include <bits/stdc++.h> 2 using namespace std; 3 4 #define ll long long 5 #define N 200010 6 int n, q, x, a[N]; 7 ll sum[2][N]; 8 9 namespace SEG 10 { 11 int lazy[N << 2], v[N << 2]; 12 void pushdown(int id, int l, int r, int mid) 13 { 14 if (lazy[id] == -1) return; 15 lazy[id << 1] = lazy[id]; 16 lazy[id << 1 | 1] = lazy[id]; 17 v[id << 1] = lazy[id] * (mid - l + 1); 18 v[id << 1 | 1] = lazy[id] * (r - mid); 19 lazy[id] = -1; 20 } 21 void pushup(int id) { v[id] = v[id << 1] + v[id << 1 | 1]; } 22 void build(int id, int l, int r) 23 { 24 lazy[id] = -1, v[id] = 0; 25 if (l == r) 26 { 27 v[id] = a[l] > x; 28 return; 29 } 30 int mid = (l + r) >> 1; 31 build(id << 1, l, mid); 32 build(id << 1 | 1, mid + 1, r); 33 pushup(id); 34 } 35 void update(int id, int l, int r, int ql, int qr, int val) 36 { 37 if (l >= ql && r <= qr) 38 { 39 lazy[id] = val; 40 v[id] = val * (r - l + 1); 41 return; 42 } 43 int mid = (l + r) >> 1; 44 pushdown(id, l, r, mid); 45 if (ql <= mid) update(id << 1, l, mid, ql, qr, val); 46 if (qr > mid) update(id << 1 | 1, mid + 1, r, ql, qr, val); 47 pushup(id); 48 } 49 int query(int id, int l, int r, int ql, int qr) 50 { 51 if (r < l) return 0; 52 if (l >= ql && r <= qr) return v[id]; 53 int mid = (l + r) >> 1; 54 pushdown(id, l, r, mid); 55 int res = 0; 56 if (ql <= mid) res += query(id << 1, l, mid, ql, qr); 57 if (qr > mid) res += query(id << 1 | 1, mid + 1, r, ql, qr); 58 return res; 59 } 60 } 61 62 ll que(int r) 63 { 64 if (r < 1) return 0; 65 int a = SEG::query(1, 1, n, 1, r); 66 int b = r - a; 67 //cout << a << " " << b << endl; 68 //cout << sum[1][a] << " " << sum[0][b] << endl; 69 return (a ? sum[1][a] : 0) + (b ? sum[0][b] : 0); 70 } 71 72 int main() 73 { 74 while (scanf("%d%d%d", &n, &q, &x) != EOF) 75 { 76 sum[0][0] = 0, sum[1][0] = 0; 77 for (int i = 1; i <= n; ++i) 78 { 79 scanf("%d", a + i); 80 if (a[i] <= x) sum[0][++sum[0][0]] = a[i]; 81 else sum[1][++sum[1][0]] = a[i]; 82 } 83 for (int i = 2; i <= n; ++i) for (int j = 0; j < 2; ++j) sum[j][i] += sum[j][i - 1]; 84 SEG::build(1, 1, n); 85 for (int qq = 1, op, l, r; qq <= q; ++qq) 86 { 87 scanf("%d%d%d", &op, &l, &r); 88 if (op == 1) printf("%lld ", que(r) - que(l - 1)); 89 else if (op == 2) 90 { 91 int a = SEG::query(1, 1, n, l, r); 92 int b = (r - l + 1) - a; 93 SEG::update(1, 1, n, l, l + b - 1, 0); 94 SEG::update(1, 1, n, l + b, r, 1); 95 } 96 else 97 { 98 int a = SEG::query(1, 1, n, l, r); 99 int b = (r - l + 1) - a; 100 SEG::update(1, 1, n, l, l + a - 1, 1); 101 SEG::update(1, 1, n, l + a, r, 0); 102 } 103 } 104 } 105 return 0; 106 }
J：Special Judge
Solved.
$枚举每两条边， 判一下即可$
1 #include<bits/stdc++.h> 2 3 using namespace std; 4 5 const double eps = 1e-9; 6 const int maxn = 1e4 + 10; 7 8 int sgn(__int128 x) 9 { 10 if(x == 0) return 0; 11 else return x > 0 ? 1 : -1; 12 } 13 14 struct Point{ 15 __int128 x, y; 16 Point(){} 17 Point(__int128 _x, __int128 _y) 18 { 19 x = _x; 20 y = _y; 21 } 22 23 bool operator == (const Point &b) const 24 { 25 return sgn(x - b.x) == 0 && sgn(y - b.y) == 0; 26 } 27 28 bool operator < (const Point &b) const 29 { 30 return sgn(x - b.x) == 0 ? sgn(y - b.y) : sgn(x - b.x); 31 } 32 33 Point operator - (const Point &b) const 34 { 35 return Point(x - b.x, y - b.y); 36 } 37 38 __int128 operator ^ (const Point &b) const 39 { 40 return x * b.y - y * b.x; 41 } 42 43 __int128 operator * (const Point &b) const 44 { 45 return x * b.x + y * b.y; 46 } 47 48 }P[maxn]; 49 50 struct Line{ 51 Point s, e; 52 Line(){} 53 Line(Point _s, Point _e) 54 { 55 s = _s; 56 e = _e; 57 } 58 59 void adjust() 60 { 61 if(e < s) swap(s, e); 62 } 63 64 int segcrossseg(Line v) 65 { 66 int d1 = sgn((e - s) ^ (v.s - s)); 67 int d2 = sgn((e - s) ^ (v.e - s)); 68 int d3 = sgn((v.e - v.s) ^ (s - v.s)); 69 int d4 = sgn((v.e - v.s) ^ (e - v.s)); 70 if((d1 ^ d2) == -2 && (d3 ^ d4) == -2) return 2; 71 return (d1 == 0 && sgn((v.s - s) * (v.s - e)) <= 0) 72 || (d2 == 0 && sgn((v.e - s) * (v.e - e)) <= 0) 73 || (d3 == 0 && sgn((s - v.s) * (s - v.e)) <= 0) 74 || (d4 == 0 && sgn((e - v.s) * (e - v.e)) <= 0); 75 } 76 77 bool pointtoseg(Point p) 78 { 79 return sgn((p - s) ^ (e - s)) == 0 && sgn((p - s) * (p - e)) <= 0; 80 } 81 }L[maxn]; 82 83 int n, m; 84 int u[maxn], v[maxn]; 85 86 int main() 87 { 88 while(~scanf("%d %d", &n, &m)) 89 { 90 for(int i = 1; i <= m; ++i) scanf("%d %d", u + i, v + i); 91 for(int i = 1; i <= n; ++i) 92 { 93 int x, y; 94 scanf("%d %d", &x ,&y); 95 P[i] = Point(x, y); 96 } 97 for(int i = 1; i <= m; ++i) 98 { 99 L[i] = Line(P[u[i]], P[v[i]]); 100 L[i].adjust(); 101 } 102 int ans = 0; 103 for(int i = 1; i <= m; ++i) for(int j = i + 1; j <= m; ++j) 104 { 105 if(L[i].segcrossseg(L[j]) == 2) ans++; 106 else if(L[i].segcrossseg(L[j]) == 1) 107 { 108 if(u[i] == u[j]) 109 { 110 if(!(L[i].pointtoseg(P[v[j]]) || (L[j].pointtoseg(P[v[i]])))) continue; 111 } 112 113 if(u[i] == v[j]) 114 { 115 if(!(L[i].pointtoseg(P[u[j]]) || (L[j].pointtoseg(P[v[i]])))) continue; 116 } 117 118 if(v[i] == u[j]) 119 { 120 if(!(L[i].pointtoseg(P[v[j]]) || (L[j].pointtoseg(P[u[i]])))) continue; 121 } 122 123 if(v[i] == v[j]) 124 { 125 if(!(L[i].pointtoseg(P[u[j]]) || (L[j].pointtoseg(P[u[i]])))) continue; 126 } 127 128 ans++; 129 } 130 } 131 printf("%d ", ans); 132 } 133 return 0; 134 }