HDU1024(最大M子段和)
Max Sum Plus Plus
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 21540 Accepted Submission(s): 7215
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. ^_^
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.
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)
dp[i][j]表示前j个数组成的序列最大i子段和,则状态转移方程为
dp[i][j]=max{dp[i][j-1]+a[j],max{dp[i-1][t]}+a[j]} i-1=<t<j-1
优化1:本题数据范围太大, 所以可以用滚动数组优化
优化2:用last数组记录上一层循环得到的最大值,具体见代码
/* ID: LinKArftc PROG: 1024.cpp LANG: C++ */ #include <map> #include <set> #include <cmath> #include <stack> #include <queue> #include <vector> #include <cstdio> #include <string> #include <climits> #include <utility> #include <cstdlib> #include <cstring> #include <iostream> #include <algorithm> using namespace std; #define eps 1e-8 #define randin srand((unsigned int)time(NULL)) #define input freopen("input.txt","r",stdin) #define debug(s) cout << "s = " << s << endl; #define outstars cout << "*************" << endl; const double PI = acos(-1.0); const double e = exp(1.0); const int inf = 0x3f3f3f3f; const int INF = 0x7fffffff; typedef long long ll; const int maxn = 1000010; int num[maxn], dp[maxn], last[maxn]; int main() { int n, m; while (~scanf("%d %d", &m, &n)) { memset(dp, 0, sizeof(dp)); memset(last, 0, sizeof(last)); for (int i = 1; i <= n; i ++) scanf("%d", &num[i]); int ma; for (int i = 1; i <= m; i ++) {//分成i段 ma = INT_MIN; for (int j = i; j <= n; j ++) {//前j个数 dp[j] = max(dp[j-1], last[j-1]) + num[j];//dp[j]表示前j个划分成i段最大值,last[j-1]表示前j-1个划分成i-1段最大值 last[j-1] = ma; ma = max(ma, dp[j]); } last[n] = ma; } printf("%d\n", ma); } return 0; }
Read The Fucking Source Code ☺!