dp的两个不错的题

C - Cheapest Palindrome

Keeping track of all the cows can be a tricky task so Farmer John has installed a system to automate it. He has installed on each cow an electronic ID tag that the system will read as the cows pass by a scanner. Each ID tag's contents are currently a single string with length M (1 ≤ M ≤ 2,000) characters drawn from an alphabet of N (1 ≤ N ≤ 26) different symbols (namely, the lower-case roman alphabet).

Cows, being the mischievous creatures they are, sometimes try to spoof the system by walking backwards. While a cow whose ID is "abcba" would read the same no matter which direction the she walks, a cow with the ID "abcb" can potentially register as two different IDs ("abcb" and "bcba").

FJ would like to change the cows's ID tags so they read the same no matter which direction the cow walks by. For example, "abcb" can be changed by adding "a" at the end to form "abcba" so that the ID is palindromic (reads the same forwards and backwards). Some other ways to change the ID to be palindromic are include adding the three letters "bcb" to the begining to yield the ID "bcbabcb" or removing the letter "a" to yield the ID "bcb". One can add or remove characters at any location in the string yielding a string longer or shorter than the original string.

Unfortunately as the ID tags are electronic, each character insertion or deletion has a cost (0 ≤ cost ≤ 10,000) which varies depending on exactly which character value to be added or deleted. Given the content of a cow's ID tag and the cost of inserting or deleting each of the alphabet's characters, find the minimum cost to change the ID tag so it satisfies FJ's requirements. An empty ID tag is considered to satisfy the requirements of reading the same forward and backward. Only letters with associated costs can be added to a string.

Input

Line 1: Two space-separated integers: N and M 
Line 2: This line contains exactly M characters which constitute the initial ID string 
Lines 3.. N+2: Each line contains three space-separated entities: a character of the input alphabet and two integers which are respectively the cost of adding and deleting that character.

Output

Line 1: A single line with a single integer that is the minimum cost to change the given name tag.

Sample Input

3 4
abcb
a 1000 1100
b 350 700
c 200 800

Sample Output

900

Hint

If we insert an "a" on the end to get "abcba", the cost would be 1000. If we delete the "a" on the beginning to get "bcb", the cost would be 1100. If we insert "bcb" at the begining of the string, the cost would be 350 + 200 + 350 = 900, which is the minimum.

这个你先要明白删除一个字符和删除一个字符的操作是一样的，全部理解为删除好了

我们用dp[i][j]表示将i~j位置的字符串变为回文串的最低耗费。

可得以下递推关系：

当str[i]==str[j]时：d[i][j]=d[i+1][j-1]

前一个状态肯定是回文

 d[i][j] = min{ d[i+1][j]+value[i], d[i][j-1]+value[j] } ;    

#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <string.h>
using namespace std;
char s[];
int dp[][];
int a[];
int main() {
    int n,m;
    while(cin>>n>>m){
        getchar();
        scanf("%s",s+);
        memset(dp,,sizeof(dp));
        for(int i=;i<n;i++){
            getchar();
            char c=getchar();
            int e,f;
            cin>>e>>f;
            a[(int)c]=min(e,f);
        }
        for(int i=m;i>;i--)
        for(int j=i+;j<=m;j++){
            if(s[i]==s[j])dp[i][j]=dp[i+][j-];
            else dp[i][j]=min(dp[i+][j]+a[(int)s[i]],dp[i][j-]+a[(int)s[j]]);
        }
        cout<<dp[][m]<<endl;

    }
    return ;
}

D - A Mini Locomotive

A train has a locomotive that pulls the train with its many passenger coaches. If the locomotive breaks down, there is no way to pull the train. Therefore, the office of railroads decided to distribute three mini locomotives to each station. A mini locomotive can pull only a few passenger coaches. If a locomotive breaks down, three mini locomotives cannot pull all passenger coaches. So, the office of railroads made a decision as follows:

1. Set the number of maximum passenger coaches a mini locomotive can pull, and a mini locomotive will not pull over the number. The number is same for all three locomotives. 
2. With three mini locomotives, let them transport the maximum number of passengers to destination. The office already knew the number of passengers in each passenger coach, and no passengers are allowed to move between coaches. 
3. Each mini locomotive pulls consecutive passenger coaches. Right after the locomotive, passenger coaches have numbers starting from 1.

For example, assume there are 7 passenger coaches, and one mini locomotive can pull a maximum of 2 passenger coaches. The number of passengers in the passenger coaches, in order from 1 to 7, is 35, 40, 50, 10, 30, 45, and 60.

If three mini locomotives pull passenger coaches 1-2, 3-4, and 6-7, they can transport 240 passengers. In this example, three mini locomotives cannot transport more than 240 passengers.

Given the number of passenger coaches, the number of passengers in each passenger coach, and the maximum number of passenger coaches which can be pulled by a mini locomotive, write a program to find the maximum number of passengers which can be transported by the three mini locomotives.

Input

The first line of the input contains a single integer t (1 <= t <= 11), the number of test cases, followed by the input data for each test case. The input for each test case will be as follows: 
The first line of the input file contains the number of passenger coaches, which will not exceed 50,000. The second line contains a list of space separated integers giving the number of passengers in each coach, such that the i ^th number of in this line is the number of passengers in coach i. No coach holds more than 100 passengers. The third line contains the maximum number of passenger coaches which can be pulled by a single mini locomotive. This number will not exceed 1/3 of the number of passenger coaches. 

Output

There should be one line per test case, containing the maximum number of passengers which can be transported by the three mini locomotives.

Sample Input

1
7
35 40 50 10 30 45 60
2

Sample Output

240

一个数列，n个数，找三个k个连续数的子数列，使其和最大。

#include <stdio.h>
#include <string.h>
#include <algorithm>
using  namespace std;
int f[][],a[];
int main() {
    int t;
    scanf("%d",&t);
    while(t--) {
        int n;
        scanf("%d",&n);
        a[]=;
        for(int i=; i<=n; i++) {
            scanf("%d",&a[i]);
            a[i]+=a[i-];
        }
        int m;
        scanf("%d",&m);
        memset(f,,sizeof(f));
        for(int i=; i<=n; i++)
            for(int j=; j<; j++) {
                int k=i-m;
                if(k<) k=;
                f[i][j]=max(f[i-][j],f[k][j-]+a[i]-a[k]);
            }
        printf("%d\n",f[n][]);
    }
    return ;
}