uva 10382 Watering Grass
2013-08-31 10:39
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题意:求最少的圆能完全覆盖要求的矩形,其实一个圆有用的部分是与矩形相交的点所构成的矩形,因为多出来的部分还是要其他圆来填满,所以我们就建一个结构体保存圆所形成矩形的左,右坐标,然后就是贪心啦,求已覆盖范围内,最远的矩形添进来
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; const int MAXN = 10005; double len,wid,x[MAXN],r[MAXN]; int n,cnt; struct node { double l; double r; }arr[MAXN]; bool cmp(node a,node b) { return a.l < b.l; } int main() { while (scanf("%d",&n) != EOF) { scanf("%lf%lf",&len,&wid); cnt = 0; for (int i = 1; i <= n; i++) { cnt++; scanf("%lf%lf",&x[cnt],&r[cnt]); if (r[cnt]*2 <= wid) cnt--; } n = cnt; for (int i = 1; i <= n; i++) { double d = sqrt(r[i]*r[i]-wid*wid*0.25); arr[i].l = x[i] - d; arr[i].r = x[i] + d; } sort(arr+1,arr+1+n,cmp); if (arr[1].l > 0.0) printf("-1\n"); else { int ans = 0, cur = 1; double end = 0.0,maxr = -1.0; while (cur <= n) { while (arr[cur].l <= end && cur <= n) //找到小于当前最远的最靠右边的矩形 { maxr =max(maxr,arr[cur].r); cur++; } end = maxr; ans++; if (end > len) break; if (arr[cur].l > end) //如果有一个猛的跳到end外,就结束了 break; } if (end < len) printf("-1\n"); else printf("%d\n",ans); } } return 0; }
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