标签:getchar get bit struct can name names ras c++
利用几何关系转化以后,变成经典的区间覆盖问题
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 10010;
const double eps = 1e-11;
int n, L, w;
int p[maxn], r[maxn];
struct Seg{
double l, r;
bool operator < (const Seg &x) const{
return l < x.l || (l == x.l && r < x.r);
}
}a[maxn];
ll read(){ ll s = 0, f = 1; char ch = getchar(); while(ch < ‘0‘ || ch > ‘9‘){ if(ch == ‘-‘) f = -1; ch = getchar(); } while(ch >= ‘0‘ && ch <= ‘9‘){ s = s * 10 + ch - ‘0‘; ch = getchar(); } return s * f; }
int main(){
while(scanf("%d%d%d", &n, &L, &w) == 3){
double W = (double)w / 2.0;
int cnt = 0;
for(int i = 1 ; i <= n ; ++i){
scanf("%d%d", &p[i], &r[i]);
if(r[i] * 2 <= w) continue;
double l = sqrt((double)r[i] * r[i] - W * W);
if(p[i] - l > (double) L + eps) continue;
a[++cnt].l = p[i] - l, a[cnt].r = p[i] + l;
}
// for(int i = 1 ; i <= cnt ; ++i){
// printf("%.3lf %.3lf\n", a[i].l, a[i].r);
// }
sort(a + 1, a + 1 + cnt);
double lt = 0.0, rt = 0.0;
int ans = 0;
for(int i = 1 ; i <= cnt ;){
if(a[i].l > lt + eps){
ans = -1;
break;
}
while(a[i].l <= lt && i <= cnt){
rt = max(rt, a[i].r);
++i;
}
++ans;
lt = rt;
if(lt - L > eps) break;
}
if(rt + eps < L) ans = -1;
printf("%d\n", ans);
}
return 0;
}
UVa 10382 Watering Grass (贪心 区间覆盖)
标签:getchar get bit struct can name names ras c++
原文地址:https://www.cnblogs.com/tuchen/p/14843088.html