CERC 2004 / UVa 1335 Beijing Guards (二分&贪心&想法题)

1335 - Beijing Guards

Time limit: 3.000 seconds

http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&category=456&page=show_problem&problem=4081

Beijing was once surrounded by four rings of city walls: the Forbidden City Wall, the Imperial City Wall, the Inner City Wall, and finally the Outer City Wall. Most of these walls were demolished in the 50s and 60s to make way for roads. The walls were protected by guard towers, and there was a guard living in each tower. The wall can be considered to be a large ring, where every guard tower has exaetly two neighbors.

The guard had to keep an eye on his section of the wall all day, so he had to stay in the tower. This is a very boring job, thus it is important to keep the guards motivated. The best way to motivate a guard is to give him lots of awards. There are several different types of awards that can be given: the Distinguished Service Award, the Nicest Uniform Award, the Master Guard Award, the Superior Eyesight Award, etc. The Central Department of City Guards determined how many awards have to be given to each of the guards. An award can be given to more than one guard. However, you have to pay attention to one thing: you should not give the same award to two neighbors, since a guard cannot be proud of his award if his neighbor already has this award. The task is to write a program that determines how many different types of awards are required to keep all the guards motivated.

Input

The input contains several blocks of test eases. Each case begins with a line containing a single integerln100000, the number of guard towers. The next n lines correspond to then guards: each line contains an integer, the number of awards the guard requires. Each guard requires at least 1, and at most l00000 awards. Guardi and i + 1 are neighbors, they cannot receive the same award. The first guard and the last guard are also neighbors.

The input is terminated by a block with n = 0.

Output

For each test case, you have to output a line containing a single integer, the minimum numberx of award types that allows us to motivate the guards. That is, if we havex types of awards, then we can give as many awards to each guard as he requires, and we can do it in such a way that the same type of award is not given to neighboring guards. A guard can receive only one award from each type.

Sample Input

3
4
2
2
5
2
2
2
2
2
5
1
1
1
1
1
0

Sample Output

8
5
3

思路：
1 当n为偶数的时候，ans = max(r[i]+r[i+1])，可以自己画个图验证;
2 当n为奇数的时候就不满足了，那么我们可以利用二分答案然后判断从而求出最小值。假设有p种礼物，那么设第一个人的礼物为1～r1，那么不难发现最有的分配的策略一定是这样的：除1以外的其他人，编号为偶数的人尽量往前取，编号为奇数的人尽量往后取。这样我们只要最后判断第n个人是否与第一个人有冲突即可。
3 由于题目并没有要我们输出实际的方案，那么我们只需要记录每个人在[1,r1]和[r1+1,n]这个范围内取了几个，最后判断第n个人是否有取[1,r1]范围内的数即可。
4 注意对于n = 1的情况需要特判。

完整代码：

/*0.059s*/

#include<bits/stdc++.h>
using namespace std;
const int maxn = 100005;

int n, r[maxn], ll[maxn], rr[maxn];

/// 测试p个礼物是否足够。
/// ll[i]是第i个人拿到的“左边的礼物”总数，rr[i]类似
bool test(int p)
{
	int x = r[1], y = p - r[1];
	ll[1] = x, rr[1] = 0;
	for (int i = 2; i <= n; i++)
	{
		if (i & 1)
		{
			///因为要满足二者，所以要取最小值
			rr[i] = min(y - rr[i - 1], r[i]); /// 尽量拿右边的礼物
			ll[i] = r[i] - rr[i];
		}
		else
		{
			ll[i] = min(x - ll[i - 1], r[i]); /// 尽量拿左边的礼物
			rr[i] = r[i] - ll[i];
		}
	}
	return ll[n] == 0;
}

int main()
{
	while (scanf("%d", &n), n)
	{
		for (int i = 1; i <= n; i++) scanf("%d", &r[i]);
		if (n == 1)  /// 特判n=1
		{
			printf("%d\n", r[1]);
			continue;
		}
		r[n + 1] = r[1];
		int L = 0, R = 0;
		for (int i = 1; i <= n; i++) L = max(L, r[i] + r[i + 1]);///最小范围
		if (n & 1)
		{
			for (int i = 1; i <= n; i++) R = max(R, r[i] * 3);///注意3倍最大范围
			while (L < R)
			{
				int M = L + (R - L) / 2;
				if (test(M)) R = M;
				else L = M + 1;
			}
		}
		printf("%d\n", L);
	}
	return 0;
}