HDU1024-Max Sum Plus Plus(dp)

题目链接

http://acm.hdu.edu.cn/showproblem.php?pid=1024

题意：

给定一个序列A，长度为n，从中找出m段，使其和最大，其中n <= 1000000，m > 0

思路：

d[i][j]表示在将第i个数选入第j段的最大和，则目标max(d[k][m])，其中k ∈ [m, n];
对A[i]，可以和A[i - 1]放入同一段，也可以开辟新的一段，于是转移方程为:

d[i][j] = max(d]i -1][j], d[k][j - 1]) + a[i]; (j - 1 <= k <= i - 1);

状态数：nm， 转移：n，时间复杂度O(m*n^2)，明显会T，并且也会MLE
对d[k][j - 1]为前i - 1项的最大和，于是可以单独用一个数组pre[]保存，每次循环的时候更新，以及循环的时候注意循环顺序
并且对d[i][j]和d[i][j - 1]，第二项j可以直接省去，于是：

d[i] = max(d[i - 1], pre[i - 1]) + a[i];

代码：

#include <iostream>#include <cstring>#include <stack>#include <vector>#include <set>#include <map>#include <cmath>#include <queue>#include <sstream>#include <iomanip>#include <fstream>#include <cstdio>#include <cstdlib>#include <climits>#include <deque>#include <bitset>#include <algorithm>using namespace std;#define PI acos(-1.0)#define LL long long#define PII pair<int, int>#define PLL pair<LL, LL>#define mp make_pair#define IN freopen("in.txt", "r", stdin)#define OUT freopen("out.txt", "wb", stdout)#define scan(x) scanf("%d", &x)#define scan2(x, y) scanf("%d%d", &x, &y)#define scan3(x, y, z) scanf("%d%d%d", &x, &y, &z)#define sqr(x) (x) * (x)#define pr(x) cout << #x << " = " << x << endl#define lc o << 1#define rc o << 1 | 1#define pl() cout << endlconst int maxn = 1000000 + 5;int a[maxn], n, m;LL d[maxn], pre[maxn];void dp() {    memset(d, 0, sizeof(d));    memset(pre, 0, sizeof(pre));    LL res;    for (int j = 1; j <= m; j++) {        res = -0x3e3e3e3e;        for (int i = j; i <= n; i++) {            d[i] = max(d[i - 1], pre[i - 1]) + a[i];            pre[i - 1] = res;            res = max(res, d[i]);        }    }    printf("%lld\n", res);}int main() {    while (~scan(m)) {        scan(n);        for (int i = 1; i <= n; i++) scan(a[i]);        dp();    }    return 0;}