UVA 1335 - Beijing Guards(贪心)

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 nextnlines correspond to thenguards: each line contains an integer, the number of awards the guard requires. Each guard requires at least 1, and at most l00000 awards. Guardiandi+ 1are neighbors, they cannot receive the same award. The first guard and the last guard are also neighbors.

The input is terminated by a block withn= 0.

Output

For each test case, you have to output a line containing a single integer, the minimum numberxof award types that allows us to motivate the guards. That is, if we havextypes 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

题意：给定n个守卫战成一圈，每个守卫要r个礼物，然后问需要几种礼物分配给他们才能使得相邻两个守卫礼物都不同，

思路：二分+贪心，二分礼物数，贪心:如果是偶数个，答案为max(ri, ri + 1),如果是奇数，先把第一个人放好，然后后面每个人奇数个尽量取前面，偶数个尽量取后面，然后判断最后一个人和第一个人是不是重复即可，用一个left数组和right数组来表示以r【0】为分界线两边取的礼物数，这样一来只要最后一个人的left为0就是可以的。

代码:

#include <stdio.h>
#include <string.h>
#define max(a,b) (a)>(b)?(a):(b)
#define min(a,b) (a)<(b)?(a):(b)
const int N = 100005;


int n, r[N], L, R, left[N], right[N];


void init () {
    L = R = 0;
    for (int i = 0; i < n; i ++)
	scanf("%d", &r[i]);
    for (int i = 0; i < n - 1; i ++)
	L = max(r[i] + r[i + 1], L);
    L = max(r[n - 1] + r[0], L);
}


bool judge(int mid) {
    memset(left, 0, sizeof(left));
    memset(right, 0, sizeof(right));
    if (r[0] > mid) return false;
    left[0] = r[0]; right[0] = 0; int x = r[0], y = mid - r[0], i;
    for (i = 1; i < n; i ++) {
	if (i % 2) {
	    left[i] = min(x - left[i - 1], r[i]);
	    right[i] = r[i] - left[i];
	}
	else {
	    right[i] = min(y - right[i - 1], r[i]);
	    left[i] = r[i] - right[i];
	}
    }
    return left[n - 1] == 0;
}


int solve() {
    if (n == 1)
	return r[0];
    if (n % 2) {
	for (int i = 0; i < n; i ++)
	    R = max(r[i] * 3, R);
	while (L < R) {
	    int mid = (L + R)>>1;
	    if (judge(mid)) R = mid;
	    else L = mid + 1;
	}
	return R;
    }
    return L;
}


int main() {
    while (~scanf("%d", &n) && n) {
	init();
	printf("%d\n", solve());
    }
    return 0;
}