Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13462 Accepted Submission(s): 4425
Problem Description
Now I think you have got an AC inIgnatius.L's "Max Sum" problem. To be a brave ACMer, we alwayschallenge ourselves to more difficult problems. Now you are faced with a moredifficult problem.
Given a consecutive number sequence S1, S2,S3, S4 ... Sx, ... Sn (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ Sx ≤ 32767). We define a function sum(i, j) = Si + ... + Sj (1 ≤ i ≤ j ≤ n).
Now given an integer m (m > 0), yourtask is to find m pairs of i and j which make sum(i1, j1) + sum(i2, j2) +sum(i3, j3) + ... + sum(im, jm) maximal (ix ≤ iy ≤ jx or ix ≤ jy ≤ jx is not allowed).
But I`m lazy, I don't want to write aspecial-judge module, so you don't have to output m pairs of i and j, justoutput the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead.^_^
Input
Each test case will begin with two integersm and n, followed by n integers S1, S2, S3 ... Sn.
Process to the end of file.
Output
Output the maximal summation describedabove in one line.
Sample Input
1 3 1 2 3
2 6 -1 4 -2 3 -2 3
Sample Output
6
8
解題代碼：
//學習c++編程，STL queue，stack，sort函數
#include<stdio.h>
#include<algorithm>
using namespace std;
inta[1000000],b1[1000000]={0},b2[1000000]={0};//注意：一般全局的，靜態的變量所允許分配的最大空間比局部的，動態的變量的要大。
int main()
{
intm,n,i,j,p,q,max;
while(scanf("%d%d",&m,&n)!=EOF)
{
for(i=1;i<=n;i++)
{
scanf("%d",&a[i]);
b1[i]=b2[i]=0;
}
max=0;//對每個事例初始化max
for(i=1;i<=m;i++)
{
for(j=0;j<=n;j++)
{
if(j==0)b2[j]=0;
elseif(i<=j)
{
p=max+a[j];//*max_element(b1+i-1,b1+j)+a[j];也可以不用max改爲這一種算法但是會超時,在計算的時候把相應max求出來
q=b2[j-1]+a[j];//沒有必要再一次重複的求max了，大大節約了時間
b2[j]=p>q?p:q;
max=max>b1[j]?max:b1[j];//賦值下一個b2[j]的max
}
}
for(j=0;j<=n;j++)b1[j]=b2[j];
max=b1[i];//賦值下一行的第一個(i==j時)b2[j]的max
}
printf("%d\n",*max_element(b2+m,b2+n+1));
}
return0;
}
1.這道題是最大字串的推廣和深入，是繼最長公共子序列之後的有一個用二維數組解決的經典的問題。
2.解題思路:設b(i,j)表示數組a的前j項中i個字段和的最大值，且第i個子段含a[j](1<=i<=m,i<=j<=n)。則所求最優值顯然爲。與最大子段和問題類似地，計算b(i,j),的遞歸式爲：
其中表示第i個子段含a[j-1],而表示第i個子段僅含a[j]。初始時，b(0,j)=0,(1jn);b(i,0)=0,(1im)。
3.，活用數學方法證明，牢記遞歸式。