01背包（变种）

http://acm.hdu.edu.cn/showproblem.php?pid=2476

String painter

Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7666    Accepted Submission(s): 3738

Problem Description

There are two strings A and B with equal length. Both strings are made up of lower case letters. Now you have a powerful string painter. With the help of the painter, you can change a segment of characters of a string to any other character you want. That is, after using the painter, the segment is made up of only one kind of character. Now your task is to change A to B using string painter. What’s the minimum number of operations?

 

Input

Input contains multiple cases. Each case consists of two lines:
The first line contains string A.
The second line contains string B.
The length of both strings will not be greater than 100.

 

Output

A single line contains one integer representing the answer.

 

Sample Input

zzzzzfzzzzz abcdefedcba abababababab cdcdcdcdcdcd

 

Sample Output

6 7

 

Source

2008 Asia Regional Chengdu

 

Recommend

lcy

题意：有n头牛，每一头牛有两个值，一个聪明值，一个幽默值。求所以牛两值之和的最大值，且其中聪明值之和和幽默值之和不能为负数

解法：可以发现此题问题就是01背包的取与不取的问题，但是这题背包容量会出现负数（负数不能作为数组下标）所以要将负数将为正数。

 

//#include <bits/stdc++.h>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <algorithm>
#include <iostream>
#include <cstdio>
#include <string>
#include <cstring>
#include <stdio.h>
#include <queue>
#include <stack>;
#include <map>
#include <set>
#include <string.h>
#include <vector>
#define ME(x , y) memset(x , y , sizeof(x))
#define SF(n) scanf("%d" , &n)
#define rep(i , n) for(int i = 0 ; i < n ; i ++)
#define INF  0x3f3f3f3f
#define mod 998244353
#define PI acos(-1)
using namespace std;
typedef long long ll ;
int dp[200005];
int w[209] , val[209];


int main()
{
    int n ;
    while(~scanf("%d" , &n))
    {
        for(int i = 0 ; i <= 200000 ; i++)
            dp[i] = -INF ;
        dp[100000] = 0 ;
        for(int i = 1 ; i <= n ; i++)
        {
            scanf("%d%d" , &w[i] , &val[i]);
        }
        for(int i = 1 ; i <= n ; i++)
        {
            if(w[i] < 0 && val[i] < 0)
                continue ;
            if(w[i] > 0)
            {
                for(int j = 200000 ; j >= w[i] ; j--)
                {
                    if(dp[j-w[i]] > -INF)
                    dp[j] = max(dp[j] , dp[j-w[i]]+val[i]);
                }
            }
            else
            {
                for(int j = w[i] ; j <= 200000 + w[i] ; j++)
                {
                    if(dp[j-w[i]] > -INF)
                    dp[j] = max(dp[j] , dp[j-w[i]]+val[i]);
                }
            }
        }
        int ans = -INF ;
        for(int i = 100000 ; i <= 200000 ; i++)
        {
            if(dp[i] >= 0)
            {
                ans = max(ans , dp[i]+i-100000);
            }
        }
        cout << ans << endl;
    }


    return 0 ;
}