Codeforces All-Ukrainian School Olympiad in Informatics F.Tourist
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题意:
有一个旅行者在一维坐标做移动,每秒可以向左或向右移不超过v个单位,在这个坐标轴上时刻ti在xi会有一个事件发生,问旅行者从0出发或从任意点出发最多能经历事件数
题解:
将每个事件看做二维坐标上的点(xi,yi)显然从事件j能赶往事件i,需满足|xi-xj| <= (ti-tj)*v,则满足改条件的区域在坐标系上呈三角形分布,条件等价于-xi + ti*v >= -xj + tj*v && xi + ti*v >= xj + tj*v,那么用新坐标
(-xi + ti*v,xi + ti*v) 替换所有点,显然i能赶往j,那么i在j左下角,控制x递增,然后就是求y坐标的最长不下降序列。第一问只操作第一象限点即可
应该是使用upper_bound而不是lower_bound.......
#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn = 1E5 + 10;const int INF = ~0U>>1;typedef long long LL;struct data{int p,q;bool operator < (const data &b) const {if (p < b.p) return 1;if (p > b.p) return 0;return q < b.q;}}a[maxn];int d[maxn],p[maxn],q[maxn];int n,v,x[maxn],t[maxn],f[maxn];int main(){#ifdef YZYfreopen("yzy.txt","r",stdin);#endifcin >> n;for (int i = 1; i <= n; i++) scanf("%d %d",&x[i],&t[i]),d[i] = INF; d[0] = -d[1];cin >> v;for (int i = 1; i <= n; i++) a[i].p = x[i] + t[i]*v,a[i].q = -x[i] + t[i]*v;sort(a + 1,a + n + 1);for (int i = 1; i <= n; i++) {int po = upper_bound(d,d + n + 1,a[i].q) - d;f[i] = po;d[po] = a[i].q;}int ans2 = 0;for (int i = 1; i <= n; i++) ans2 = max(ans2,f[i]);for (int i = 1; i <= n+1; i++) d[i] = INF; d[0] = 0;memset(f,0,sizeof(f));for (int i = 1; i <= n; i++) {if (a[i].p < 0 || a[i].q < 0) continue;int po = upper_bound(d,d + n + 1,a[i].q) - d;f[i] = po;d[po] = a[i].q;} int ans1 = 0;for (int i = 1; i <= n; i++) ans1 = max(ans1,f[i]);printf("%d %d",ans1,ans2);return 0;}
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