201C Fragile Bridges

dp，参考了http://www.cppblog.com/hanfei19910905/archive/2012/06/30/180831.html；

定义两组状态 L1[i] 表示在 0-i 内的最大得分，L2[i] 表示从 i 出发回到 i 的最大得分，不难得到状态转移方程，最后枚举 i 组合求出最大值；

# include <cstdio>
# include <algorithm>

# define N 100005

using namespace std;

typedef long long int LL;

int n, num[N];
LL L1[N], L2[N], R1[N], R2[N], ans;

void init(void)
{
    int i;

    scanf("%d", &n);
    for (i = 1; i < n; ++i)
        scanf("%d", &num[i]);
}

void solve(void)
{
    int i;

    L1[0] = L2[0] = 0;
    for (i = 1; i < n; ++i)
    {
        L2[i] = (num[i]>1 ?  L2[i-1]+num[i]/2*2:0);
        if (num[i] & 0x1) L1[i] = L1[i-1] + num[i];
        else L1[i] = max(L2[i], L1[i-1]+num[i]-1);
    }
    R2[n-1] = R1[n-1] = 0;
    for (i = n-2; i >= 0; --i)
    {
        R2[i] = (num[i+1]>1 ? R2[i+1]+num[i+1]/2*2:0);
        if (num[i+1] & 0x1) R1[i] = R1[i+1] + num[i+1];
        else R1[i] = max(R2[i], R1[i+1]+num[i+1]-1);
    }
    ans = 0;
    for (i = 0; i < n; ++i)
    {
        ans = max(ans, max(max(L1[i], R1[i]), max(L1[i]+R2[i], L2[i]+R1[i])));
    }
    printf("%I64d\n", ans);
}

int main()
{
    //freopen("in.txt", "r", stdin);

    init();
    solve();

    return 0;
}