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;

}