总Time Limit: 1000msMemory Limit: 65536kB
Description
Given a set of n integers: A={a1, a2,..., an}, we define a function d(A) as below: t1 t2
d(A) = max{ ∑ai + ∑aj | 1 <= s1 <= t1 < s2 <= t2 <= n } i=s1 j=s2
Your task is to calculate d(A).
Input
The input consists of T(<=30) test cases. The number of test cases (T) is given in the first line of the input.
Each test case contains two lines. The first line is an integer n(2<=n<=50000). The second line contains n integers: a1, a2, ..., an. (|ai| <= 10000).There is an empty line after each case.
Output
Print exactly one line for each test case. The line should contain the integer d(A).
Sample Input
1
10
1 -1 2 2 3 -3 4 -4 5 -5
Sample Output
13
Hint
In the sample, we choose {2,2,3,-3,4} and {5}, then we can get the answer Huge input,scanf is recommended.
求两块最大的互不相交的字序列的和
f[i]表示以i结尾的最大子序列的和
ff[i]表示i之前的最大子序列的和
g[i]表示以i开头的最大子序列的和
gg[i]表示i之后的最大子序列的和
max=ff[i]+gg[i+1];
code：
var t:longint;init:array[1..50000]of longint; ii,i,j,k,n:longint; f,g,ff,gg:array[1..50000]of longint; max:longint; max1,max2:longint; function findmax(x,y:longint):longint; begin if x>y then exit(x) else exit(y); end; begin readln(t); for ii:=1 to t do begin readln(n); for i:=1 to n do read(init[i]); f[1]:=init[1]; ff[1]:=f[1]; max1:=f[1]; for i:=2 to n do begin f[i]:=findmax(f[i-1],0)+init[i]; if f[i]>max1 then max1:=f[i]; ff[i]:=max1; end; g[n]:=init[n]; gg[n]:=init[n]; max2:=gg[n]; for i:=n-1 downto 1 do begin g[i]:=findmax(g[i+1],0)+init[i]; if g[i]>max2 then max2:=g[i]; gg[i]:=max2; end; max:=-maxlongint; for i:=1 to n-1 do if max<ff[i]+gg[i+1] then max:=ff[i]+gg[i+1]; writeln(max); end; end.
