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

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

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

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

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

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

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