EC-final2017 C - Traffic Light Gym - 101775C（读题）

One year ago, Mr. Ang joined a great company and he rented a house on the same street as the company. The company is so great that Mr. Ang doesn’t need to punch in and out. He can have a good sleep and gets up at any time.

Every day, he walks along the street, goes through N traffic lights numbered 1, 2, 3, …, N and arrives the company. It takes Mr. Ang S0 seconds from the house to first traffic light. Si seconds from traffic light i to traffic light i + 1 and SN seconds from Nth traffic light to company. The time spent on the way to company depends on the state of traffic lights.

Mr. Ang got interested in the traffic lights and hacked into the system. After reading the code, he found that for traffic light i on his way, it stays Ai seconds green and then stays Bi seconds red and repeats. He also found that for all traffic lights, Ai + Bi are the same. He can modify the code to set different offsets OFFi for the traffic lights. Formally, Mr. Ang can set the offsets for the traffic lights to OFFi so that for each traffic light, Mr. Ang can go through it in time range [OFFi + k × (Ai + Bi), OFFi + k × (Ai + Bi) + Ai) and must wait in time range [OFFi + k × (Ai + Bi) + Ai, OFFi + (k + 1) × (Ai + Bi)) for all integers k.

Now Mr. Ang would like to make his commuting time from house to company as short as possible in the worst case. Find out the minimal time in second.

Input
The first line of the input gives the number of test cases, T. T test cases follow.

Each test case starts with a line which consists of one number N, indicating the number of traffic lights.

The next line consists of N + 1 numbers S0, S1, …, SN indicating the walking time between house, traffic lights and company in seconds as described in the problem.

Then followed by N lines each consists of 2 numbers Ai, Bi indicating the length of green light and red light of the ith traffic light.

1 ≤ T ≤ 50.
1 ≤ N ≤ 1000.
1 ≤ Ai, Bi ≤ 120. All Ai + Bi are the same.
1 ≤ Si ≤ 106.
Output
For each test case, output one line containing “Case #x: y”, where x is the test case number (starting from 1) and y is the minimal time in second from house to company in the worst case.

y will be considered correct if it is within an absolute or relative error of 10 - 8 of the correct answer.

Example
Input
2
1
30 70
15 15
2
30 15 70
10 20
20 10
Output
Case #1: 115.000000
Case #2: 135.000000
Note
In the first test case, it takes Mr. Ang 30 seconds to the first traffic light, in the worst case he had to wait 15 seconds until it gets green. Then 70 seconds to the company. Total time is 30 + 15 + 70 = 115 seconds;

In the second test case, it still takes Mr. Ang 30 seconds to the first traffic light, as Mr. Ang could make OFF1 = 0, OFF2 = 25. If Mr. Ang meets red light at the first traffic light, he will definitely meet green light at the seconds one. So in the worst case, it takes Mr. And 30 + 20 + 15 + 0 + 70 = 135 seconds.

题意懂了就能a…

题意：
有n个红绿灯。给出绿灯红灯维持的时间。
你可以设置每个红绿灯的延迟时间，就是本来k1时间是绿灯，k2时间是红灯，变成k1+off时间是绿灯，k2 + off时间是红灯，其他顺推。
你可以在任意时间出发。
求出每种off设置方式的最大时间的最小值。

思路：
可以看到，无论你怎么设置延迟时间，最坏情况总会通过最长红灯。同时可以构造出通过最长红灯但是其他都是绿灯的最坏情况。此时这个最坏情况花费最小。
所以加上最大红灯时间即可。

ACNEW

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <vector>
#include <queue>

using namespace std;

typedef long long ll;

int main() {
    int T;scanf("%d",&T);
    int kase = 0;
    while(T--) {
        int n;scanf("%d",&n);
        double sum = 0;
        for(int i = 1;i <= n + 1;i++) {
            ll x;scanf("%lld",&x);
            sum += x;
        }
        ll mx = 0;
        for(int i = 1;i <= n;i++) {
            ll x,y;scanf("%lld%lld",&x,&y);
            mx = max(mx,y);
        }
        sum += mx;
        printf("Case #%d: %.8f\n",++kase,sum);
    }
    return 0;
}

队友code

#include <queue>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

const int maxn = 1e3 + 7;
const int mod = 1000000007;
typedef long long ll;

int t, n;
double num[maxn], a[maxn], b[maxn];

int main() {
	scanf("%d", &t); 
	for (int cse = 1; cse <= t; ++cse) {
		scanf("%d", &n); double mx = 0;
		for (int i = 1; i <= n + 1; ++i) {
			double x; scanf("%lf", &x);
			num[i] = num[i - 1] + x;
		}
		for (int i = 1; i <= n; ++i) {
			scanf("%lf%lf", &a[i], &b[i]);
			mx = max(mx, b[i]);
		}
		printf("Case #%d: %.8f\n", cse, num[n + 1] + mx);
	}
	return 0;
}