[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。
1 #pragma comment(linker, "/STACK:10240000,10240000") 2 3 #include <iostream> 4 #include <cstdio> 5 #include <algorithm> 6 #include <cstdlib> 7 #include <cstring> 8 #include <map> 9 #include <queue> 10 #include <deque> 11 #include <cmath> 12 #include <vector> 13 #include <ctime> 14 #include <cctype> 15 #include <set> 16 17 using namespace std; 18 19 #define mem0(a) memset(a, 0, sizeof(a)) 20 #define lson l, m, rt << 1 21 #define rson m + 1, r, rt << 1 | 1 22 #define define_m int m = (l + r) >> 1 23 #define rep(a, b) for(int a = 0; a < b; a++) 24 #define rrep(a, b) for(int a = b - 1; a >= 0; a--) 25 #define all(a) (a).begin(), (a).end() 26 #define lowbit(x) ((x) & (-(x))) 27 #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) {} 28 #define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {} 29 #define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {} 30 #define pc(a) putchar(a) 31 #define ps(a) printf("%s", a) 32 #define pd(a) printf("%d", a) 33 #define sd(a) scanf("%d", &a) 34 35 typedef double db; 36 typedef long long LL; 37 typedef pair<int, int> pii; 38 typedef multiset<int> msi; 39 typedef set<int> si; 40 typedef vector<int> vi; 41 typedef map<int, int> mii; 42 43 const int dx[8] = {0, 1, 0, -1, 1, 1, -1, -1}; 44 const int dy[8] = {1, 0, -1, 0, -1, 1, 1, -1}; 45 const int maxn = 1e5 + 7; 46 const int maxm = 1e5 + 7; 47 const int maxv = 1e7 + 7; 48 const int max_val = 1e6 + 7; 49 const int MD = 1e9 +7; 50 const int INF = 1e9 + 7; 51 const double PI = acos(-1.0); 52 const double eps = 1e-10; 53 54 template<class T> T gcd(T a, T b) { return b == 0? a : gcd(b, a % b); } 55 56 LL sum[maxn], sum2[maxn]; 57 int a[maxn]; 58 int n, p, m ,t, x; 59 60 LL work() { 61 LL ans = 0; 62 int l = max(0, p - t), r = min(n - 1, p + t); 63 for (int i = l; i < p; i++) { 64 int pos1 = p + min(m / 2, p - i), pos2 = min(m + i, p + p - i), pos = max(pos1, pos2 + t - (p - i)); 65 pos = min(r, pos); 66 ans = max(ans, sum[pos] - (i? sum[i - 1] : 0)); 67 } 68 ans = max(ans, sum[r] - (p? sum[p - 1] : 0)); 69 return ans; 70 } 71 72 int main() { 73 //freopen("in.txt", "r", stdin); 74 75 while (cin >> n >> p) { 76 p--; 77 sum[0] = 0; 78 rep(i, n) { 79 sd(a[i]); 80 if (i) sum[i] = sum[i - 1]; 81 sum[i] += a[i]; 82 } 83 sum2[0] = a[n - 1]; 84 rrep(i, n - 1) sum2[n - i - 1] = sum2[n - i - 2] + a[i]; 85 cin >> m >> t; 86 87 LL ans = work(); 88 memcpy(sum, sum2, sizeof(sum)); 89 p = n - p - 1; 90 ans = max(ans, work()); 91 cout << ans << endl; 92 } 93 return 0; 94 }
