SOJ 4479 Easy Problem III(区间贪心)

题意:

$给定一条无限长的直线,给定N\le 10^5条线段[s, t]覆盖这条直线$
$问覆盖的长度(重复覆盖只算一次)$

分析:

$经典区间贪心问题,左端点排序,然后维护最右端点就好了$
$累加答案的时候,记得判断大小关系。。。$
$时间复杂度O(nlogn)$

代码:

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//
// Created by TaoSama on 2016-04-09
// Copyright (c) 2016 TaoSama. All rights reserved.
//
#pragma comment(linker, "/STACK:102400000,102400000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>
using namespace std;
#define pr(x) cout << #x << " = " << x << " "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;
int n;
pair<int, int> a[N];
int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
// freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);
int t; scanf("%d", &t);
while(t--) {
scanf("%d", &n);
for(int i = 1; i <= n; ++i) scanf("%d%d", &a[i].first, &a[i].second);
sort(a + 1, a + 1 + n);
int ans = a[1].second - a[1].first, r = a[1].second;
for(int i = 2; i <= n; ++i) {
if(a[i].first > r) ans += a[i].second - a[i].first;
else if(a[i].second > r) ans += a[i].second - r;
r = max(r, a[i].second);
}
printf("%d\n", ans);
}
return 0;
}


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