USACO-Section1.3 Milking Cows (区间问题)
2017-05-30 16:32
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2017-5-30
题目描述
解答
代码
或者这么写…其实是一样一样的。
按理来说应该是可以按照开始时间排序然后求解的。
题目描述
给你几个区间,求出最长的连续长度以及最短的连续长度
解答
最长为1000000,在区间内则赋值为1,否则不操作,最后计算出结果 即可,这种方法耗时比较长,注意区间的开闭问题
代码
/* ID: 18795871 PROG: milk2 LANG: C++ */ #include<iostream> #include<fstream> #include<cstring> using namespace std; const int N = 1000000; bool f[N+1]; ifstream fin("milk2.in"); ofstream fout("milk2.out"); void res(int a,int b){ for (int i=a;i<b;i++) f[i]=true; } int main() { int i,n,a,b,ma=-N,mi=N; fin>>n; memset(f,false,sizeof(f)); for (i=0;i<n;i++){ fin>>a>>b; ma=max(a,ma);ma=max(ma,b); mi=min(a,mi);mi=min(mi,b); res(a,b); } int res1=-N,res2=-N,sum1,sum2;i=mi; while (i<ma){ sum1=0;sum2=0; while (!f[i]&&i<=ma){ sum1++;i++; } while (f[i]&&i<=ma){ sum2++;i++; } res1=max(res1,sum1); res2=max(res2,sum2); } fout<<res2<<" "<<res1<<endl; return 0; }
或者这么写…其实是一样一样的。
/* ID: 18795871 PROG: milk2 LANG: C++ */ #include<iostream> #include<fstream> #include<cstring> using namespace std; ifstream fin("milk2.in"); ofstream fout("milk2.out"); const int N = 1000000; bool f[N+1]; int n; int main(){ int i,j,e,s,mis,mae; while (fin>>n){ mis=N;mae=0; memset(f,false,sizeof(f)); for (i=0;i<n;i++){ fin>>s>>e; mis=min(mis,s); mae=max(mae,e); for (j=s;j<e;j++){ f[j]=1; } } int l=0,h=1,r1=0,r2=0; for (i=mis+1;i<=mae;i++){ if (f[i]==f[i-1]){ if (f[i]) h++; else l++; }else{ r1=max(r1,l); r2=max(r2,h); if (f[i]){ l=0;h=1; }else{ l=1;h=0; } } } fout<<r2<<" "<<r1<<endl; } return 0; }
按理来说应该是可以按照开始时间排序然后求解的。
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