Max Sum Plus Plus HDU - 1024（动态规划求最大M子段和）
題意：
----最大M子段和問題
給定由 n個整數（可能為負整數）組成的序列以及一個正整數 m，要求確定序列的 m個不相交子段，使這m個子段的總和達到最大，求出最大和。
題目：
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. _
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.
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
思路：
很明顯，我們不能知道某一點是否該加入段中，加入到那一段中，不能用貪心寫，那么只能用動態規劃，看某點j是否需要加入i段中，或是新開一段作為新段的開頭。
AC代碼
#include<stdio.h> #include<string.h> #include<algorithm> using namespace std; typedef long long ll; const int M=1e6+10; const int inf=0x3f3f3f3f; int n,m,dp[M],f[M],s[M],ans; int main() {while(~scanf("%d%d",&n,&m)){for(int i=1; i<=m; i++)scanf("%d",&f[i]);memset(dp,0,sizeof(dp));memset(s,0,sizeof(s));for(int i=1; i<=n; i++){ans=-inf;for(int j=i; j<=m; j++){dp[j]=max(dp[j-1]+f[j]/*f[j]屬于第i段*/,s[j-1]+f[j]/*f[j]不屬于第i段為新的段加上前i-1段，看是否較大*/);/*看該點是否為新段的開始*/s[j-1]=ans;/*上一段（i-1)到j-1點的最大值*/ans=max(dp[j],ans);/*在第i段第j點的最大值，為當前最大值*/}}printf("%d\n",ans);}return 0; }總結
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