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这题0ms应该怎么做?我的贪心过程如下:
num=0;
for(i=0;i<n;){
num++;
for(j=i+1;j<n;j++) //如果j区间的左边小于或等于i区的右边,则此区间可以覆盖
if(check(p[j].l,p[i].r)!=1) continue;
else break;
i=j; //i跳转到下一个需要新增radar的区间
}
p的l是区间的左边,r是区间的右边
o(n)时间,为什么还不能0ms??
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