UVA 1335 Beijing Guards

Input
The input contains several blocks of test eases. Each case begins with a line containing a single integer
l ≤ n ≤ 100000, the number of guard towers. The next n lines correspond to the n guards: each line
contains an integer, the number of awards the guard requires. Each guard requires at least 1, and at
most l00000 awards. Guard i and i + 1 are neighbors, they cannot receive the same award. The rst
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 number x of
award types that allows us to motivate the guards. That is, if we have x 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

分奇偶两种情况，奇要二分，思考相应的策略，画图思考

#include <cstdio>
#include <algorithm>

using namespace std;

int A[100000 + 10], n, p;

bool ok(int p) {
    int x = A[0], y = p - x, l1 = x, r1 = y;
    for (int i = 1, l, r; i < n; ++i, l1 = l, r1 = r) {
        if (i & 1) {
            l = min(x - l1, A[i]); //放左边
            r = A[i] - l;
        } else {
            r = min(y - r1, A[i]); //放右边
            l = A[i] - r;
        }
    }

    return !l1;  //左边不应有值
}


int main() {
    while (scanf("%d", &n) && n) {
        for (int i = 0; i < n; ++i) {
            scanf("%d", &A[i]);
        }
        if (p = A[0], n == 1) goto la;
        A[n] = A[0];
        for (int i = 1; i <= n; ++i) {
            p = max(A[i] + A[i - 1], p);
        }

        if (n & 1) {
            int l = p, r = 0;
            for (int i = 0; i < n; ++i) {
                r = max(r, 3 * A[i]);
            }
            while (l < r) {
                int m = (l + r) >> 1;
                if (ok(m))
                    r = m;
                else
                    l = m + 1;
            }
            p = r;
        }
        la:
        printf("%d\n", p);
    }
}