区间&背包dp练习题题解（Ａ B C T）

A题

Gappu has a very busy weekend ahead of him. Because, next weekend is Halloween, and he is planning to attend as many parties as he can. Since it’s Halloween, these parties are all costume parties, Gappu always selects his costumes in such a way that it blends with his friends, that is, when he is attending the party, arranged by his comic-book-fan friends, he will go with the costume of Superman, but when the party is arranged contest-buddies, he would go with the costume of ‘Chinese Postman’.

Since he is going to attend a number of parties on the Halloween night, and wear costumes accordingly, he will be changing his costumes a number of times. So, to make things a little easier, he may put on costumes one over another (that is he may wear the uniform for the postman, over the superman costume). Before each party he can take off some of the costumes, or wear a new one. That is, if he is wearing the Postman uniform over the Superman costume, and wants to go to a party in Superman costume, he can take off the Postman uniform, or he can wear a new Superman uniform. But, keep in mind that, Gappu doesn’t like to wear dresses without cleaning them first, so, after taking off the Postman uniform, he cannot use that again in the Halloween night, if he needs the Postman costume again, he will have to use a new one. He can take off any number of costumes, and if he takes off k of the costumes, that will be the last k ones (e.g. if he wears costume A before costume B, to take off A, first he has to remove B).

Given the parties and the costumes, find the minimum number of costumes Gappu will need in the Halloween night.

Input
Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case starts with a line containing an integer N (1 ≤ N ≤ 100) denoting the number of parties. Next line contains N integers, where the ith integer ci (1 ≤ ci ≤ 100) denotes the costume he will be wearing in party i. He will attend party 1 first, then party 2, and so on.

Output
For each case, print the case number and the minimum number of required costumes.

Sample Input
2
4
1 2 1 2
7
1 2 1 1 3 2 1

Sample Output
Case 1: 3
Case 2: 4

这是老师讲的一道区间dp的例题
题意：
给你n天需要穿的衣服的样式，每次可以套着穿衣服，脱掉的衣服就不能再用了（可以再穿），问至少要带多少条衣服才能参加所有宴会

思路：区间DP
dp[i][j]代表从区间i到区间j需要的最少穿衣服数量。
考虑第i天，如果后面的[i+1, j]天的衣服没有和第i天相同的，那么dp[i][j] = dp[i + 1][j] + 1。
然后在区间[i +1, j]里面找到和第i天衣服一样的日子，尝试直到那天都不把i脱掉，也就是说如果这件衣服不脱 掉的话，
那么就会有dp[i][j] = dp[i + 1][k - 1] + dp[k][j]，取所有可能情况的一个最小值就可以。

代码

#include 
#include 
#include 
#include

using namespace std;
typedef long long LL;
const int N = 1e2 + 10;
int n, m;
int num[N];
int dp[N][N];

int main()
{
    int t;
    cin >> t;
    int Case = 0;
    while (t--)
    {
        cin >> n;
        for (int i = 1; i <= n; ++i)
            cin>>num[i];
        for (int i = n; i >= 1; --i)
            for (int j = i; j <= n; ++j)
            {
                dp[i][j] = dp[i + 1][j] + 1;
                for (int k = i + 1; k <= j; ++k)
                {
                    if (num[k] == num[i])dp[i][j] = min(dp[i][j], dp[i + 1][k - 1] + dp[k][j]);
                }
            }
        printf("Case %d: %d\n", Case++, dp[1][n]);
    }
}

B题

We give the following inductive definition of a “regular brackets” sequence:

the empty sequence is a regular brackets sequence,
if s is a regular brackets sequence, then (s) and [s] are regular brackets sequences, and
if a and b are regular brackets sequences, then ab is a regular brackets sequence.
no other sequence is a regular brackets sequence
For instance, all of the following character sequences are regular brackets sequences:

(), [], (()), ()[], ()[()]

while the following character sequences are not:

(, ], )(, ([)], ([(]

Given a brackets sequence of characters a1a2 … an, your goal is to find the length of the longest regular brackets sequence that is a subsequence of s. That is, you wish to find the largest m such that for indices i1, i2, …, im where 1 ≤ i1 < i2 < … < im ≤ n, ai1ai2 … aim is a regular brackets sequence.

Given the initial sequence ([([]])], the longest regular brackets subsequence is [([])].

Input
The input test file will contain multiple test cases. Each input test case consists of a single line containing only the characters (, ), [, and ]; each input test will have length between 1 and 100, inclusive. The end-of-file is marked by a line containing the word “end” and should not be processed.

Output
For each input case, the program should print the length of the longest possible regular brackets subsequence on a single line.

Sample Input
((()))
()()()
([]])
)[)(
([][][)
end

Sample Output
6
6
4
0
6

这个题也是老师讲的区间DP的例题

题意：
给出一个 只有()[]的 字符串，问最多有多少括号能匹配（一对括号算2）

思路：区间DP
看起来很像经典的栈应用，但是那个是判断是否完全匹配的

dp[i][j]为区间[i,j]的最大完全匹配数
那么对于dp[i][j]
1.如果边界两个匹配，显然有 dp[i][j]=dp[i+1][j-1]+2;
2.如果边界不匹配，那么dp[i][j] = max(dp[i][j] , dp[i][k] + dp[k+1][j])。

代码

#include
#include
#include
#include
using namespace std;
const int N = 100 + 10;
int dp[N][N];
char str[N];
int main() 
{
	while (cin>>str + 1&&str[1] != 'e') 
	{
		memset(dp, 0, sizeof(dp));
		int n = strlen(str + 1);
		for (int len = 2; len <= n; len++) 
		{
			for (int i = 1; i <= n; i++) 
			{
				int j = i + len - 1;
				if (j > n) 
					continue;
				if (str[i] == '(' && str[j] == ')' || str[i] == '[' && str[j] == ']') 
					dp[i][j] = dp[i + 1][j - 1] + 2;
				for (int k = i; k < j; k++) 
					dp[i][j] = max(dp[i][j], dp[i][k] + dp[k + 1][j]);
			}
		}
		printf("%d\n", dp[1][n]);
	}
	return 0;
}

C题

The multiplication puzzle is played with a row of cards, each containing a single positive integer. During the move player takes one card out of the row and scores the number of points equal to the product of the number on the card taken and the numbers on the cards on the left and on the right of it. It is not allowed to take out the first and the last card in the row. After the final move, only two cards are left in the row.

The goal is to take cards in such order as to minimize the total number of scored points.

For example, if cards in the row contain numbers 10 1 50 20 5, player might take a card with 1, then 20 and 50, scoring
10150 + 50205 + 10505 = 500+5000+2500 = 8000

If he would take the cards in the opposite order, i.e. 50, then 20, then 1, the score would be
15020 + 1205 + 1015 = 1000+100+50 = 1150.

Input
The first line of the input contains the number of cards N (3 <= N <= 100). The second line contains N integers in the range from 1 to 100, separated by spaces.

Output
Output must contain a single integer - the minimal score.

Sample Input
6
10 1 50 50 20 5

Sample Output
3650

老师讲过的例题

题意：
给出一堆数字，每次可以移除一个数字（首尾数字不能移除），移除一个数字可以得到的分数为相邻两个数字以及他本身这三个数的乘积，最终会剩下首尾两个数字。
问可以得到的最小分数为多少？

思路

定义状态 dp[i][j] ：将第i个至第j个中间的数字全部取走可以得到的最小分数
枚举在区间(i,j)最后一个被取走的数字为第k个，则有转移方程
dp[i][j]=min(dp[i][j],dp[i][k]+dp[k][j]+a[i]∗a[k]∗a[j])

代码：

#include
#include
#include
#include
using namespace std;
const int maxn = 110;
int n, a[maxn], dp[maxn][maxn];

int main()
{
    while (scanf("%d", &n) != EOF)
    {
        for (int i = 1; i <= n; i++) 
            scanf("%d", &a[i]);
        memset(dp, 0x3f, sizeof(dp));
        for (int l = 2; l < n; l++)
        {
            for (int i = 1; i <= n - l; i++)
            {
                int j = i + l;
                for (int k = i + 1; k <= j - 1; k++)
                {
                    int tmp = a[i] * a[j] * a[k];
                    if (k > i + 1) tmp += dp[i][k];
                    if (j > k + 1) tmp += dp[k][j];
                    dp[i][j] = min(dp[i][j], tmp);
                }
            }
        }
        printf("%d\n", dp[1][n]);
    }
    return 0;
}

T题

Before ACM can do anything, a budget must be prepared and the necessary financial support obtained. The main income for this action comes from Irreversibly Bound Money (IBM). The idea behind is simple. Whenever some ACM member has any small money, he takes all the coins and throws them into a piggy-bank. You know that this process is irreversible, the coins cannot be removed without breaking the pig. After a sufficiently long time, there should be enough cash in the piggy-bank to pay everything that needs to be paid.

But there is a big problem with piggy-banks. It is not possible to determine how much money is inside. So we might break the pig into pieces only to find out that there is not enough money. Clearly, we want to avoid this unpleasant situation. The only possibility is to weigh the piggy-bank and try to guess how many coins are inside. Assume that we are able to determine the weight of the pig exactly and that we know the weights of all coins of a given currency. Then there is some minimum amount of money in the piggy-bank that we can guarantee. Your task is to find out this worst case and determine the minimum amount of cash inside the piggy-bank. We need your help. No more prematurely broken pigs!

Input
The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing two integers E and F. They indicate the weight of an empty pig and of the pig filled with coins. Both weights are given in grams. No pig will weigh more than 10 kg, that means 1 <= E <= F <= 10000. On the second line of each test case, there is an integer number N (1 <= N <= 500) that gives the number of various coins used in the given currency. Following this are exactly N lines, each specifying one coin type. These lines contain two integers each, Pand W (1 <= P <= 50000, 1 <= W <=10000). P is the value of the coin in monetary units, W is it’s weight in grams.
Output

Print exactly one line of output for each test case. The line must contain the sentence “The minimum amount of money in the piggy-bank is X.” where X is the minimum amount of money that can be achieved using coins with the given total weight. If the weight cannot be reached exactly, print a line “This is impossible.”.

Sample Input
3
10 110
2
1 1
30 50
10 110
2
1 1
50 30
1 6
2
10 3
20 4

Sample Output
The minimum amount of money in the piggy-bank is 60.
The minimum amount of money in the piggy-bank is 100.
This is impossible.

老师讲的例题

思路：
很基本的完全背包问题，直接写就ok，就是注意是求最小

代码

#include 
#include 
#include 
#include
using namespace std;
const int MAX_N = 1000003;
int v[MAX_N], w[MAX_N], dp[MAX_N];
int main() 
{
    int n, E, F, N, i, j, W;
    cin >> n;
    while (n--) 
    {
        cin >> E >> F;
        cin >> N;
        W = F - E;
        memset(v, 0, sizeof(v));
        memset(w, 0, sizeof(w));
        for (i = 1; i <= W; i++) 
            dp[i] = MAX_N;
        dp[0] = 0;
        for (i = 1; i <= N; i++) 
            cin >> v[i] >> w[i];
        for (i = 1; i <= N; i++)
            for (j = w[i]; j <= W; j++) 
                dp[j] = min(dp[j], dp[j - w[i]] + v[i]);
        if (dp[W] != MAX_N) 
            cout << "The minimum amount of money in the piggy-bank is " << dp[W] << "." << endl;
        else 
            cout << "This is impossible." << endl;
    }
    return 0;
}