USACO-Section1.2 Milking Cows [其他]

2017-05-21
题目大意：

给定N个区间，有重叠部分的区间可以合并成一个区间，求合并后最长的区间长度，和最大的区间间隙（即两个区间之间的间隔）。

题解：

用pair <int, int>存储所有区间，按照first升序排序，然后合并有重叠部分的区间。再在合并后的区间里找最大的区间长度和区间间隙。

代码：

/*ID: zachery1PROG: milk2LANG: C++*/#include <iostream>#include <fstream>#include <algorithm>#include <vector>#define cin fin#define cout foutusing namespace std;ifstream fin("milk2.in");ofstream fout("milk2.out");typedef pair<int, int> P;vector<P> line;vector<P> mergedLine;int N;void merge() {    int begin = line[0].first, end = line[0].second;    for (int i = 1; i < N; i++) {        if (line[i].first <= end) {            if (line[i].second <= end) continue;            else {                end = line[i].second;            }        }        else {            mergedLine.push_back(make_pair(begin, end));            begin = line[i].first;            end = line[i].second;        }    }    mergedLine.push_back(make_pair(begin, end));}void solve() {    int maxMilking, maxIdle;    maxMilking = mergedLine[0].second - mergedLine[0].first;    maxIdle = 0;    for (int i = 1; i < mergedLine.size(); i++) {        maxMilking = max(maxMilking, mergedLine[i].second - mergedLine[i].first);        maxIdle = max(maxIdle, mergedLine[i].first - mergedLine[i-1].second);    }    cout << maxMilking << " " << maxIdle << endl;}int main() {    cin >> N;    for (int i = 0; i < N; i++) {        int begin, end;        cin >> begin >> end;        line.push_back(make_pair(begin, end));    }       sort(line.begin(), line.end());    merge();    solve();    return 0;}