HDU 1231 /POJ 2479 简单DP之最大字段和

原题地址:http://acm.hdu.edu.cn/showproblem.php?pid=1231
算法核心:
动态规划, 数组为 $vec[]$ ，设 $dp[i]$ 是以 $vec[i]$ 结尾的子数组的最大和，对于元素 $vec[i+1],$ 它有两种选择： $a、vec[i+1]$ 接着前面的子数组构成最大和， $b$ 、 $vec[i+1]$ 自己单独构成子数组。则 $dp[i+1] = max{dp[i]+vec[i+1], vec[i+1]}$

#include <bits/stdc++.h>
#define eps 1e-8
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,(rt<<1)+1
#define CLR(x,y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
int n;
int a[maxn];
int main() {
    while (~scanf("%d", &n), n) {
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        int sum, sumstart, maxsum, maxstart, maxend;
        sum = sumstart = -INF;
        maxsum = maxstart = maxend = -INF;
        for (int i = 1; i <= n; i++) {
            if (sum < 0) {
                sum = a[i];
                sumstart = a[i];
            } else sum += a[i];
            if (sum > maxsum) {
                maxsum = sum;
                maxstart = sumstart;
                maxend = a[i];
            }
        }
        if (maxsum < 0) {
            printf("0 %d %d\n", a[1], a[n]);
        } else {
            printf("%d %d %d\n", maxsum, maxstart, maxend);
        }
    }
    return 0;
}

原题地址:http://poj.org/problem?id=2479

思路: $dp1[i]$ 表示第一个元素 $a[1]$ 到 $a[i]$ 的最大子段和，状态转移方程是
$dp1[i] = max(dp1[i-1]+a[i],a[i])$
$dp2[i]$ 表示末一个元素 $a[n]$ 到 $a[i]$ 的最大子段和，状态转移方程是
$dp2[i] = max(dp2[i+1]+a[i],a[i]);$
再用 $ans1[i]$ 表示 $max({dp1[1],dp1[2],...,dp1[i]})$ ,
用 $ans2[i]$ 表示 $max({dp2[n],dp2[n-1],...,dp2[i]}),$

#include <cmath>
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <queue>
#include <vector>
#include <stack>
#include <set>
#include <map>
#include <cctype>
#define eps 1e-8
#define INF 0x3f3f3f3f
#define PI acos(-1)
#define lson l,mid,rt<<1
#define rson mid+1,r,(rt<<1)+1
#define CLR(x,y) memset((x),y,sizeof(x))
#define fuck(x) cerr << #x << "=" << x << endl

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int seed = 131;
const int maxn = 1e5 + 5;
const int mod = 998244353;
ll dp1[maxn], dp2[maxn], ans1[maxn], ans2[maxn];
ll a[maxn];
bool cmp(ll a, ll b) {
    return a > b;
}
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        int  n;
        scanf("%d", &n);
        CLR(dp1, 0);
        CLR(dp2, 0);
        CLR(ans1, 0);
        CLR(ans2, 0);
        for (int i = 1; i <= n; i++) {
            scanf("%lld", &a[i]);
        }
        dp1[1] = ans1[1] = a[1];
        dp2[n] = ans2[n] = a[n];
        for (int i = 2; i <= n; i++) {
            dp1[i] = max(dp1[i - 1] + a[i], a[i]);
            ans1[i] = max(ans1[i - 1], dp1[i]);
        }
        for (int i = n - 1; i >= 1; i--) {
            dp2[i] = max(dp2[i + 1] + a[i], a[i]);
            ans2[i] = max(ans2[i + 1], dp2[i]);
        }
        ll MAX = -1e14;
        for (int i = 2; i <= n; i++) {
            MAX = max(MAX, ans1[i - 1] + ans2[i]);
        }
        printf("%lld\n", MAX);
    }
    return 0;
}

总结一下,有两种方法

一:dp[i]表示以第i个数结尾的区间最大和,状态转移方程是dp[i] = max(dp[i-1]+a[i],a[i])

二:详见第一段代码,即下面的代码


int MaxArray( int n, int arr_[])
{
    int i, sum_ = 0, max_ = 0;
    for (i=1; i<=n; i++)
    {
        if (sum_>0)
        {
            sum_ += arr_[i];
        }
        else
        {
            sum_ = arr_[i];
        }
        if (sum_>max_)
        {
            max_ = sum_;
        }
    }
    return max_;
}

HDU 1231 /POJ 2479 简单DP之最大字段和

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