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 S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (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(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x 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(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Given a consecutive number sequence S 1, S 2, S 3, S 4 ... S x, ... S n (1 ≤ x ≤ n ≤ 1,000,000, -32768 ≤ S x ≤ 32767). We define a function sum(i, j) = S i + ... + S j (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(i 1, j 1) + sum(i 2, j 2) + sum(i 3, j 3) + ... + sum(i m, j m) maximal (i x ≤ i y ≤ j x or i x ≤ j y ≤ j x 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(i x, j x)(1 ≤ x ≤ m) instead. ^_^
Input
Each test case will begin with two integers m and n, followed by n integers S
1, S
2, S
3 ... S
n.
Process to the end of file.
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 }