[zoj3627]模拟吧

思路：情况只可能是2种，两个人一直向一边走，或者有一个人折回来，对于后一种，枚举折回来的位置就行了。不过要注意两个方向都要处理下。

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <set>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep(a, b) for(int a = 0; a < b; a++)
#define rrep(a, b) for(int a = b - 1; a >= 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pc(a) putchar(a)
#define ps(a) printf("%s", a)
#define pd(a) printf("%d", a)
#define sd(a) scanf("%d", &a)

typedef double db;
typedef long long LL;
typedef pair<int, int> pii;
typedef multiset<int> msi;
typedef set<int> si;
typedef vector<int> vi;
typedef map<int, int> mii;

const int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1};
const int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1};
const int maxn = 1e5 + 7;
const int maxm = 1e5 + 7;
const int maxv = 1e7 + 7;
const int max_val = 1e6 + 7;
const int MD = 1e9 +7;
const int INF = 1e9 + 7;
const double PI = acos(-1.0);
const double eps = 1e-10;

template<class T> T gcd(T a, T b) { return b == 0? a : gcd(b, a % b); }

LL sum[maxn], sum2[maxn];
int a[maxn];
int n, p, m ,t, x;

LL work() {
    LL ans = 0;
    int l = max(0, p - t), r = min(n - 1, p + t);
    for (int i = l; i < p; i++) {
        int pos1 = p + min(m / 2, p - i), pos2 = min(m + i, p + p - i), pos = max(pos1, pos2 + t - (p - i));
        pos = min(r, pos);
        ans = max(ans, sum[pos] - (i? sum[i - 1] : 0));
    }
    ans = max(ans, sum[r] - (p? sum[p - 1] : 0));
    return ans;
}

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

    while (cin >> n >> p) {
        p--;
        sum[0] = 0;
        rep(i, n) {
            sd(a[i]);
            if (i) sum[i] = sum[i - 1];
            sum[i] += a[i];
        }
        sum2[0] = a[n - 1];
        rrep(i, n - 1) sum2[n - i - 1] = sum2[n - i - 2] + a[i];
        cin >> m >> t;

        LL ans = work();
        memcpy(sum, sum2, sizeof(sum));
        p = n - p - 1;
        ans = max(ans, work());
        cout << ans << endl;
    }
    return 0;
}