hdu---1024Max Sum Plus Plus(动态规划)
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15898 Accepted Submission(s): 5171
Problem Description
Now I think you have got an AC in Ignatius.L's "Max Sum" problem. To be a brave ACMer, we always challenge ourselves to more difficult problems. Now you are faced with a more difficult 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), your task 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 a special-judge module, so you don't have to output m pairs of i and j, just output the maximal summation of sum(ix, jx)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S1, S2, S3 ... Sn. Process to the end of file.
Output
Output the maximal summation described above in one line.
Sample Input
1 3 1 2 3
2 6 -1 4 -2 3 -2 3
Sample Output
6
8
Hint
Huge input, scanf and dynamic programming is recommended.
Author
JGShining（极光炫影）
题意：
给定一段连续数字串，要你求出m个连续子串的最大值。
比如
下面这个样列
2 6 -1 4 -2 3 -2 3
给出一个长度为6的连续数字串，要你求出这个串中连续的m=2个子串的子串之和.......
aa | -1 | 4 | -2 | 3 | -2 | 3 | ||||
|---|---|---|---|---|---|---|---|---|---|---|
第一遍 maxc | -1 | 4 | 4 | 4 | 5 | \ | ||||
dp | -1 | 4 | 2 | 5 | 3 | 6 | ||||
第二遍 maxc | -1 | 3 | 3 | 7 | 7 | \ | ||||
dp | -1 | 3 | 2 | 7 | 5 | 8 | ||||
代码
1 #include<iostream>
2 #include<string.h>
3 #include<stdio.h>
4 using namespace std;
5 int a[1000001],dp[1000001],max1[1000001];
6 int max(int x,int y){
7 return x>y?x:y;
8 }
9 int main(){
10 int i,j,n,m,temp;
11 while(scanf("%d%d",&m,&n)!=EOF)
12 {
13 dp[0]=0;
14 for(i=1;i<=n;i++)
15 {
16 scanf("%d",&a[i]);
17 dp[i]=0;
18 max1[i]=0;
19 }
20 max1[0]=0;
21 for(i=1;i<=m;i++){
22 temp=-0x3f3f3f3f;
23 for(j=i;j<=n;j++){
24 dp[j]=max(dp[j-1]+a[j],max1[j-1]+a[j]);
25 max1[j-1]=temp;
26 temp=max(temp,dp[j]);
27 }
28 }
29 printf("%d\n",temp);
30 }
31 return 0;
32 }优化后的代码：
1 /*hdu 1024 @coder Gxjun*/
2 #include<iostream>
3 #include<cstdio>
4 #include<cstring>
5 #include<cstdlib>
6 using namespace std;
7 const int maxn=1000005;
8 int aa[maxn],dp[maxn],maxc[maxn];
9 int max(int a,int b){
10 return a>b?a:b;
11 }
12 int main()
13 {
14 int n,m,i,j,temp;
15 while(scanf("%d%d",&m,&n)!=EOF){
16 memset(maxc,0,sizeof(int)*(n+1));
17 memset(dp,0,sizeof(int)*(n+1));
18 for(i=1;i<=n;i++)
19 scanf("%d",&aa[i]);
20 for(i=1;i<=m;i++){
21 temp=-0x3f3f3f3f;
22 for(j=i;j<=n;j++){
23 dp[j]=max(dp[j-1],maxc[j-1])+aa[j];
24 maxc[j-1]=temp;
25 temp=max(temp,dp[j]);
26 }
27 }
28 printf("%d\n",temp);
29 }
30 }