思路：半夜了思路有点混乱wa了好几发。一开始坑定两个人距离为m才能获得最大的收益，所以我们就可以枚举单个端点，当距离达到m时在一同一个方向走这是我们只需要算一下剩下几秒，左右两边贪心去最大的即可。
代码如下：
1 /************************************************** 2 * Author : xiaohao Z 3 * Blog : http://www.cnblogs.com/shu-xiaohao/ 4 * Last modified : 2014-05-12 23:51 5 * Filename : poj_1741.cpp 6 * Description : 7 * ************************************************/ 8 9 #include <iostream> 10 #include <cstdio> 11 #include <cstring> 12 #include <cstdlib> 13 #include <cmath> 14 #include <algorithm> 15 #include <queue> 16 #include <stack> 17 #include <vector> 18 #include <set> 19 #include <map> 20 #define MP(a, b) make_pair(a, b) 21 #define PB(a) push_back(a) 22 23 using namespace std; 24 typedef long long ll; 25 typedef pair<int, int> pii; 26 typedef pair<unsigned int,unsigned int> puu; 27 typedef pair<int, double> pid; 28 typedef pair<ll, int> pli; 29 typedef pair<int, ll> pil; 30 31 const int INF = 0x3f3f3f3f; 32 const double eps = 1E-6; 33 const int LEN = 1000000+10; 34 ll num[LEN], sum[LEN]; 35 int n, p, m, t; 36 37 int Min(int a, int b, int c){ 38 return min(a, min(b, c)); 39 } 40 41 int Max(int a, int b, int c){ 42 return max(a, max(b, c)); 43 } 44 45 void J(int &pos){ 46 if(pos == n+1) pos = n; 47 else if(pos == 0) pos = 1; 48 } 49 50 int main() 51 { 52 // freopen("in.txt", "r", stdin); 53 54 while(scanf("%d%d", &n, &p)!=EOF){ 55 sum[0] = 0; 56 for(int i=1; i<=n; i++){ 57 scanf("%lld", &num[i]); 58 sum[i] = sum[i-1] + num[i]; 59 } 60 scanf("%d%d", &m, &t); 61 ll ans = 0; 62 for(int l=max(1, p-t); l<=p; l++){ 63 int r = Min(n, l+m, p+t); 64 int rest = t - max(p-l, r-p); 65 int a = max(1, l-rest); 66 int b = min(n, r+rest); 67 ans = max(ans, sum[b]-sum[l-1]); 68 ans = max(ans, sum[r]-sum[a-1]); 69 } 70 for(int r=min(n, p+t); r>=p; r--){ 71 int l = Max(1, r-m, p-t); 72 int rest = t - max(p-l, r-p); 73 int a = max(1, l-rest); 74 int b = min(n, r+rest); 75 ans = max(ans, sum[b]-sum[l-1]); 76 ans = max(ans, sum[r]-sum[a-1]); 77 } 78 printf("%lld ", ans); 79 } 80 return 0; 81 }