算是一个总结吧!
先来一个模板;
TYVJ 1305 最大子序和
题目描述
输入一个长度为n的整数序列,从中找出一段不超过M的连续子序列,使得整个序列的和最大。例如 1,-3,5,1,-2,3
当m=4时,S=5+1-2+3=7
当m=2或m=3时,S=5+1=6
输入输出格式
输入格式:第一行两个数n,m
第二行有n个数,要求在n个数找到最大子序和
一个数,数出他们的最大子序和
输入输出样例
6 4 1 -3 5 1 -2 3
7
数据范围:
100%满足n,m<=300000
这个可以算是单调队列优化dp的模板题了吧;
令f[i]表示以i为结尾的连续子序的最大值;
所以, f[i] = min (sum[i] - sum[j]) ( i - m <= j <= i); sum 为前缀和;
即 f[i] = sum[i] - min(sum[j])( i - m <= j <= i);
显然可以用单调队列维护前缀最小值;
代码:
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <queue> using namespace std; int n, m; int a[300010]; int sum[300010]; deque <int> q; int ans; int main() { cin >> n >> m; for(register int i = 1 ; i <= n ; i ++) { scanf("%d", &a[i]); sum[i] = sum[i-1] + a[i]; } for(register int i = 1 ; i <= n ; i ++) { while(!q.empty() && sum[q.front()] > sum[i]) q.pop_front(); q.push_front(i); while(!q.empty() && i - m > q.back()) q.pop_back(); if(i != 1) { ans = max(ans, sum[i] - sum[q.back()]); } else ans = max(ans, sum[i]); } cout << ans << endl; return 0; }
再来一道水题;
POJ1821 Fence
Description
Being the team's leader you want to determine for each worker the interval that he should paint, knowing that the total income should be maximal. The total income represents the sum of the workers personal income.
Write a program that determines the total maximal income obtained by the K workers.
Input
Input
N K
L1 P1 S1
L2 P2 S2
...
LK PK SK
Semnification
N -the number of the planks; K ? the number of the workers
Li -the maximal number of planks that can be painted by worker i
Pi -the sum received by worker i for a painted plank
Si -the plank in front of which sits the worker i
Output
Sample Input
8 4 3 2 2 3 2 3 3 3 5 1 1 7
Sample Output
17
Hint
the worker 1 paints the interval [1, 2];
the worker 2 paints the interval [3, 4];
the worker 3 paints the interval [5, 7];
the worker 4 does not paint any plank
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <queue> using namespace std; #define int long long int n, m; struct Par { int l, p, s; }a[105]; bool operator < (Par a, Par b) { return a.s < b.s; } int dp[105][16005]; deque <int> q; inline int cacl(int x, int y) { return dp[x-1][y] - a[x].p * y; } signed main() { while(scanf("%lld%lld", &n, &m) != EOF) //cin >> n >> m; { for(register int i = 1 ; i <= m ; i ++) { scanf("%lld%lld%lld", &a[i].l, &a[i].p, &a[i].s); } sort(a + 1, a + 1 +m); for(register int i = 1 ; i <= m ; i ++) { for(register int k = max((int)0, a[i].s - a[i].l) ; k <= a[i].s - 1 ; k ++) { while(!q.empty() && cacl(i, q.back()) <= cacl(i, k)) q.pop_back(); q.push_back(k); } for(register int j = 1 ; j <= n ; j ++) { dp[i][j] = max(dp[i-1][j], dp[i][j-1]); if(j >= a[i].s) { while(!q.empty() && q.front() < j - a[i].l) q.pop_front(); if(!q.empty()) dp[i][j] = max(dp[i][j], cacl(i, q.front()) + a[i].p * j); } } } cout << dp[m][n] << endl; } return 0; }
来个复杂点的:
SCOI2010 股票交易
提交地址 :股票交易
题目描述
最近lxhgww又迷上了投资股票,通过一段时间的观察和学习,他总结出了股票行情的一些规律。
通过一段时间的观察,lxhgww预测到了未来T天内某只股票的走势,第i天的股票买入价为每股APi,第i天的股票卖出价为每股BPi(数据保证对于每个i,都有APi>=BPi),但是每天不能无限制地交易,于是股票交易所规定第i天的一次买入至多只能购买ASi股,一次卖出至多只能卖出BSi股。
另外,股票交易所还制定了两个规定。为了避免大家疯狂交易,股票交易所规定在两次交易(某一天的买入或者卖出均算是一次交易)之间,至少要间隔W天,也就是说如果在第i天发生了交易,那么从第i+1天到第i+W天,均不能发生交易。同时,为了避免垄断,股票交易所还规定在任何时间,一个人的手里的股票数不能超过MaxP。
在第1天之前,lxhgww手里有一大笔钱(可以认为钱的数目无限),但是没有任何股票,当然,T天以后,lxhgww想要赚到最多的钱,聪明的程序员们,你们能帮助他吗?
输入输出格式
输入格式:
输入数据第一行包括3个整数,分别是T,MaxP,W。
接下来T行,第i行代表第i-1天的股票走势,每行4个整数,分别表示APi,BPi,ASi,BSi。
输出格式:
输出数据为一行,包括1个数字,表示lxhgww能赚到的最多的钱数。
输入输出样例
说明
对于30%的数据,0<=W<T<=50,1<=MaxP<=50
对于50%的数据,0<=W<T<=2000,1<=MaxP<=50
对于100%的数据,0<=W<T<=2000,1<=MaxP<=2000
对于所有的数据,1<=BPi<=APi<=1000,1<=ASi,BSi<=MaxP
设dp[i][j]为第i天拥有j个股票的最大收益;
则dp[i] [j] = dp[i-1][j] 这一天不买、卖股票;
dp[i] [j] = max( dp[i-w-1] [k] + bp[i] * ( j - k) ) , 这是卖出股票, 满足j - k <= bs[i] ;
同理:dp[i] [j] = max( dp[i-w-1] [k] -ap[i] * (k - j) ) 这是买入股票, 满足k - j <= as[i];
然后上面两个式子可以用单调队列优化,具体看代码;
![](https://images.cnblogs.com/OutliningIndicators/ContractedBlock.gif)
//By zZhBr #include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include <queue> using namespace std; #define maxn 2010 #define int long long int t, p, w; int ap[maxn], bp[maxn], as[maxn], bs[maxn]; int dp[maxn][maxn]; deque <int> q; inline int calc1(int i, int j) { return dp[i-w-1][j] + bp[i] * j; } inline int calc2(int i, int j) { return dp[i-w-1][j] + ap[i] * j; } signed main() { memset(dp, 0xcf, sizeof dp); cin >> t >> p >> w; for(register int i = 1 ; i <= t ; i ++) { scanf("%lld%lld%lld%lld", &ap[i], &bp[i], &as[i], &bs[i]); } dp[0][0] = 0; for(register int i = 1 ; i <= t ; i ++) { for(register int j = 0 ; j <= as[i] ; j ++) dp[i][j] = -ap[i] * j; for(register int j = p ; j >= 0 ; j --) { dp[i][j] = max(dp[i][j], dp[i-1][j]); } if(i - w - 1 >= 0) { q.clear(); for(register int j = p ; j >= 0 ; j --) { if(1)//卖出 { while(!q.empty() && calc1(i, q.back()) <= calc1(i, j)) q.pop_back(); q.push_back(j); while(!q.empty() && - j + q.front() > bs[i]) q.pop_front(); //dp[i][j] = max(dp[i][j], dp[i-w-1][k] + bp[i] * (k - j)); dp[i][j] = max(dp[i][j], calc1(i, q.front()) - bp[i] * j); } } q.clear(); for(register int j = 0 ; j <= p ; j ++) { if(1)//买入 { while(!q.empty() && calc2(i, q.back()) <= calc2(i, j)) q.pop_back(); q.push_back(j); while(!q.empty() && j - q.front() > as[i]) q.pop_front(); //dp[i][j] = max(dp[i][j], dp[i-w-1][k] - ap[i] * (j - k)); dp[i][j] = max(dp[i][j], calc2(i, q.front()) - ap[i] * j); } } } } int ans = 0; for(register int i = 0 ; i <= p ; i ++) { ans = max(ans, dp[t][i]); } cout << ans << endl; return 0; }