BZOJ - 1537: [POI2005]Aut- The Bus
题解：把车站看成点，以$x$为第一关键字，$y$为第二关键字排序。记$dp_i$为纵坐标为$y$时的最大值，转移就行。
1 #include<cstdio> 2 #include<iostream> 3 #include<algorithm> 4 using namespace std; 5 inline char nc() { 6 static char b[1<<16],*s=b,*t=b; 7 return s==t&&(t=(s=b)+fread(b,1,1<<16,stdin),s==t)?-1:*s++; 8 } 9 inline void read(int &x) { 10 char b = nc(); x = 0; 11 for (; !isdigit(b); b = nc()); 12 for (; isdigit(b); b = nc()) x = x * 10 + b - '0'; 13 } 14 struct P { 15 int x, y, z; 16 inline void init() { 17 read(x), read(y), read(z); 18 } 19 inline bool operator<(const P &p) const { 20 return x != p.x ? x < p.x : y < p.y; 21 } 22 } p[100005]; 23 int n, m, k, c[100005], H, ans, v[100005], dp[100005]; 24 inline void gmax(int &x, int y) { 25 if (x < y) x = y; 26 } 27 inline int lowbit(int x) { 28 return x & -x; 29 } 30 inline void add(int p, int x) { 31 for (; p <= H; p += lowbit(p)) gmax(c[p], x); 32 } 33 inline int mx(int p) { 34 int res = 0; 35 for (; p; p -= lowbit(p)) gmax(res, c[p]); 36 return res; 37 } 38 inline int h(int x) { 39 return lower_bound(v + 1, v + 1 + H, x) - v; 40 } 41 int main() { 42 read(n); read(m); read(k); 43 for (int i = 1; i <= k; ++i) p[i].init(), v[i] = p[i].y; 44 sort(v + 1, v + 1 + k); H = unique(v + 1, v + 1 + k) - v - 1; 45 for (int i = 1; i <= k; ++i) p[i].y = h(p[i].y); 46 sort(p + 1, p + 1 + k); 47 for (int i = 1; i <= k; ++i) { 48 dp[i] = mx(p[i].y) + p[i].z; 49 gmax(ans, dp[i]); 50 add(p[i].y, dp[i]); 51 } printf("%d\n", ans); 52 return 0; 53 }
