UVALive 6834Shopping （贪心）

来源：互联网发布：js高级程序设计最新编辑：程序博客网时间：2024/05/23 13:27

题意:

N个点,给定M个限定先后访问关系,求从0出发全部访问完毕到N+1的最短时间

分析:

画画图发现可以贪心,对于交叉的区间我们总是可以合并成一个大区间一起访问,最后得到一些不交叉的区间
总时间为n+1+每个区间的长度∗2

代码:

////  Created by TaoSama on 2015-12-04//  Copyright (c) 2015 TaoSama. All rights reserved.////#pragma comment(linker, "/STACK:1024000000,1024000000")#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 << endlconst int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;typedef pair<int, int> P;int n, m;P a[1005];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);    while(scanf("%d%d", &n, &m) == 2) {        for(int i = 1; i <= m; ++i) scanf("%d%d", &a[i].first, &a[i].second);        sort(a + 1, a + 1 + m);        int l = 0, r = 0, ans = n + 1;        for(int i = 1; i <= m; ++i) {            if(a[i].first <= r) r = max(r, a[i].second);            else {                ans += r - l << 1;                l = a[i].first, r = a[i].second;            }        }        ans += r - l << 1;        printf("%d\n", ans);    }    return 0;}

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