[USACO07DEC]观光奶牛Sightseeing Cows 二分答案+判断负环
题目描述
Farmer John has decided to reward his cows for their hard work by taking them on a tour of the big city! The cows must decide how best to spend their free time.
Fortunately, they have a detailed city map showing the L (2 ≤ L ≤ 1000) major landmarks (conveniently numbered 1.. L) and the P (2 ≤ P ≤ 5000) unidirectional cow paths that join them. Farmer John will drive the cows to a starting landmark of their choice, from which they will walk along the cow paths to a series of other landmarks, ending back at their starting landmark where Farmer John will pick them up and take them back to the farm. Because space in the city is at a premium, the cow paths are very narrow and so travel along each cow path is only allowed in one fixed direction.
While the cows may spend as much time as they like in the city, they do tend to get bored easily. Visiting each new landmark is fun, but walking between them takes time. The cows know the exact fun values Fi (1 ≤ Fi ≤ 1000) for each landmark i.
The cows also know about the cowpaths. Cowpath i connects landmark L1i to L2i (in the direction L1i -> L2i ) and requires time Ti (1 ≤ Ti ≤ 1000) to traverse.
In order to have the best possible day off, the cows want to maximize the average fun value per unit time of their trip. Of course, the landmarks are only fun the first time they are visited; the cows may pass through the landmark more than once, but they do not perceive its fun value again. Furthermore, Farmer John is making the cows visit at least two landmarks, so that they get some exercise during their day off.
Help the cows find the maximum fun value per unit time that they can achieve.
作为对奶牛们辛勤工作的回报,Farmer JohnFarmer\ JohnFarmer John决定带她们去附近的大城市玩一天。旅行的前夜,奶牛们在兴奋地讨论如何最好地享受这难得的闲暇。
很幸运地,奶牛们找到了一张详细的城市地图,上面标注了城市中所有L(2⩽L⩽1000)L(2\leqslant L\leqslant1000)L(2⩽L⩽1000)座标志性建筑物(建筑物按1…L1\dots L1…L顺次编号),以及连接这些建筑物的P(2⩽P⩽5000)P(2\leqslant P\leqslant5000)P(2⩽P⩽5000)条道路。按照计划,那天早上Farmer JohnFarmer\ JohnFarmer John会开车将奶牛们送到某个她们指定的建筑物旁边,等奶牛们完成她们的整个旅行并回到出发点后,将她们接回农场。由于大城市中总是寸土寸金,所有的道路都很窄,政府不得不把它们都设定为通行方向固定的单行道。
尽管参观那些标志性建筑物的确很有意思,但如果你认为奶牛们同样享受穿行于大城市的车流中的话,你就大错特错了。与参观景点相反,奶牛们把走路定义为无趣且令她们厌烦的活动。对于编号为iii的标志性建筑物,奶牛们清楚地知道参观它能给自己带来的乐趣值Fi(1⩽Fi⩽1000)F_i (1\leqslant F_i\leqslant1000)Fi(1⩽Fi⩽1000)。相对于奶牛们在走路上花的时间,她们参观建筑物的耗时可以忽略不计。
奶牛们同样仔细地研究过城市中的道路。她们知道第i条道路两端的建筑物L1iL1_iL1i和L2iL2_iL2i(道路方向为L1i→L2iL1_i \rightarrow L2_iL1i→L2i),以及她们从道路的一头走到另一头所需要的时间Ti(1⩽Ti⩽1000)T_i(1\leqslant T_i\leqslant1000)Ti(1⩽Ti⩽1000)。
为了最好地享受她们的休息日,奶牛们希望她们在一整天中平均每单位时间内获得的乐趣值最大。当然咯,奶牛们不会愿意把同一个建筑物参观两遍,也就是说,虽然她们可以两次经过同一个建筑物,但她们的乐趣值只会增加一次。顺便说一句,为了让奶牛们得到一些锻炼,Farmer JohnFarmer\ JohnFarmer John要求奶牛们参观至少222个建筑物。
请你写个程序,帮奶牛们计算一下她们能得到的最大平均乐趣值。
输入输出格式
输入格式:
* Line 1: Two space-separated integers: L and P
* Lines 2..L+1: Line i+1 contains a single one integer: Fi
* Lines L+2..L+P+1: Line L+i+1 describes cow path i with three space-separated integers: L1i , L2i , and Ti
输出格式:
* Line 1: A single number given to two decimal places (do not
perform explicit rounding), the maximum possible average fun per unit
time, or 0 if the cows cannot plan any trip at all in accordance with
the above rules.
输入输出样例
6.00
二分答案:
我们要求的是:
假设当前答案为 x;
如果可以更新,即:
然后将每一条边的边权设为:
用 spfa 看是否有负环即可;
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize(2)
using namespace std;
#define maxn 200005
#define inf 0x7fffffff
//#define INF 1e18
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
} ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
int sqr(int x) { return x * x; } /*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/ int n, m;
int a[maxn];
struct node {
int u, v, w;
int nxt;
}edge[maxn];
int head[maxn];
double dis[maxn];
bool vis[maxn];
int num[maxn];
int tot; void addedge(int u, int v, int w) {
edge[++tot].u = u; edge[tot].v = v; edge[tot].w = w;
edge[tot].nxt = head[u]; head[u] = tot;
}
bool check(double x) {
queue<int>q;
for (int i = 1; i <= n; i++) {
q.push(i); vis[i] = num[i] = 1;
dis[i] = 0.0;
}
while (!q.empty()) {
int u = q.front(); q.pop();
vis[u] = 0;
for (int i = head[u]; i; i = edge[i].nxt) {
int v = edge[i].v;
if (dis[v] > dis[u] + x * edge[i].w - (double)a[u]) {
dis[v] = dis[u] + x * edge[i].w - (double)a[u];
if (!vis[v]) {
q.push(v); vis[v] = 1;
num[v]++;
if (num[v] >= n)return true;
}
}
}
}
return false;
} int main() {
//ios::sync_with_stdio(0);
rdint(n); rdint(m);
for (int i = 1; i <= n; i++)rdint(a[i]);
for (int i = 1; i <= m; i++) {
int u, v, w; rdint(u); rdint(v); rdint(w);
addedge(u, v, w);
}
double l = 0, r = 1e5+1;
for (int i = 1; i <= 100; i++) {
double mid = (l + r) / 2.0;
if (check(mid)) {
l = mid;
}
else r = mid;
}
printf("%.2lf\n", 1.0*l);
return 0;
}
[USACO07DEC]观光奶牛Sightseeing Cows 二分答案+判断负环的更多相关文章
- P2868 [USACO07DEC]观光奶牛Sightseeing Cows
P2868 [USACO07DEC]观光奶牛Sightseeing Cows [](https://www.cnblogs.com/images/cnblogs_com/Tony-Double-Sky ...
- 洛谷P2868 [USACO07DEC]观光奶牛Sightseeing Cows
P2868 [USACO07DEC]观光奶牛Sightseeing Cows 题目描述 Farmer John has decided to reward his cows for their har ...
- 洛谷 P2868 [USACO07DEC]观光奶牛Sightseeing Cows
题目描述 Farmer John has decided to reward his cows for their hard work by taking them on a tour of the ...
- 洛谷P2868 [USACO07DEC]观光奶牛 Sightseeing Cows
题目描述 Farmer John has decided to reward his cows for their hard work by taking them on a tour of the ...
- POJ3621或洛谷2868 [USACO07DEC]观光奶牛Sightseeing Cows
一道\(0/1\)分数规划+负环 POJ原题链接 洛谷原题链接 显然是\(0/1\)分数规划问题. 二分答案,设二分值为\(mid\). 然后对二分进行判断,我们建立新图,没有点权,设当前有向边为\( ...
- 洛谷P2868 [USACO07DEC]观光奶牛Sightseeing Cows(01分数规划)
题意 题目链接 Sol 复习一下01分数规划 设\(a_i\)为点权,\(b_i\)为边权,我们要最大化\(\sum \frac{a_i}{b_i}\).可以二分一个答案\(k\),我们需要检查\(\ ...
- Luogu 2868 [USACO07DEC]观光奶牛Sightseeing Cows
01分数规划复习. 这东西有一个名字叫做最优比率环. 首先这个答案具有单调性,我们考虑如何检验. 设$\frac{\sum_{i = 1}^{n}F_i}{\sum_{i = 1}^{n}T_i} = ...
- 洛谷 2868 [USACO07DEC]观光奶牛Sightseeing Cows
题目戳这里 一句话题意 L个点,P条有向边,求图中最大比率环(权值(Fun)与长度(Tim)的比率最大的环). Solution 巨说这是0/1分数规划. 话说 0/1分数规划 是真的难,但貌似有一些 ...
- 洛谷 P2868 [USACO07DEC]观光奶牛Sightseeing Cows 题解
题面 这道题是一道标准的01分数规划: 但是有一些细节可以优化: 不难想到要二分一个mid然后判定图上是否存在一个环S,该环是否满足∑i=1t(Fun[vi]−mid∗Tim[ei])>0 但是 ...
随机推荐
- OpenCV——饱和度调整
参考: 闲人阿发伯的博客 // define head function #ifndef PS_ALGORITHM_H_INCLUDED #define PS_ALGORITHM_H_INCLUDED ...
- centos下安装Mysql5.7.20
0.环境 本文操作系统: CentOS 7.2.1511 x86_64MySQL 版本: 5.7.16 1.卸载系统自带的 mariadb-lib [root@centos-linux ~]# rpm ...
- NOI2018网络同步赛游记
Day1 t1是一道NOI选手眼中的送分题,对于我来说还是有难度的,用了个把小时想了出来可持久化并查集的做法,最后一个点被卡常.赛后才发现Kruskal重构树是这样的简单.t2.t3由于我真的是太弱了 ...
- 8th
2017-2018-2 20179212<网络攻防实践>第8周作业 视频学习 Kali权限维持之后门 权限维持包含Tunnel工具集.Web后门.系统后门三个子类.其中系统后门与web后门 ...
- bzoj 4817: [Sdoi2017]树点涂色 LCT+树链剖分+线段树
题目: Bob有一棵n个点的有根树,其中1号点是根节点.Bob在每个点上涂了颜色,并且每个点上的颜色不同. 定义一条路径的权值是:这条路径上的点(包括起点和终点)共有多少种不同的颜色. Bob可能会进 ...
- NOIp2018集训test-10-16 (bike day2)
“毕姥爷:今天的题好简单啊,你们怎么考得这么烂啊,如果是noip你们就凉透了啊“ 今天的题难度应该是3.2.1递减的,但是我不知道哪根筋没搭对,平时我最多1h多就弃题了,今天硬生生写了2h20min的 ...
- 白话算法(6) 散列表(Hash Table)从理论到实用(上)
处理实际问题的一般数学方法是,首先提炼出问题的本质元素,然后把它看作一个比现实无限宽广的可能性系统,这个系统中的实质关系可以通过一般化的推理来论证理解,并可归纳成一般公式,而这个一般公式适用于任何特殊 ...
- 洛谷 4149 [IOI2011]Race——点分治
题目:https://www.luogu.org/problemnew/show/P4149 第一道点分治! 点分治大约是每次找重心,以重心为根做一遍树形dp:然后对于该根的每个孩子,递归下去.递归之 ...
- 【转】Pro Android学习笔记(二三):用户界面和控制(11):其他控件
目录(?)[-] Chronometer计时器控件 倒计时CountDownTimer Switch控件 Space控件 其他控件 Android提供了很多控件,基本上都是view的扩展. Chron ...
- CSS 关于文本 背景 边框整理
文本与字体 1)阴影:text-shadow 格式:text-shadow:5px 5px 3px #FFFFFF分别对应 水平方向 垂直方向 模糊程度 颜色值 代码: <!DOCTYPE ht ...