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                Problem Description Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don't know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N (0<N<1000) kinds of facilities (different value, different kinds).
 
Input Input contains multiple test cases. Each test case starts with a number N (0 < N <= 50 -- the total number of different facilities). The next N lines contain an integer V (0<V<=50 --value of facility) and an integer M (0<M<=100 --corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.
 
Output For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.
 
Sample Input

210 120 1310 1 20 230 1-1  
Sample Output

20 1040 40  

 

題意：給出每個物體的價值和物體的數量，如何分使得A,B所得價值最接近並且A的價值不能小於B

思路：將總和平分後，就是一道01揹包題了

 

#include <stdio.h>#include <string.h>#include <algorithm>using namespace
 
 std;int val[5005];int dp[255555];int main(){    int n,i,j,a,b,l,sum;    while(~scanf("%d",&n),n>0)    {        memset(val,0,sizeof(val));        memset(dp,0,sizeof(dp));        l = 0
 
;        sum = 0;        for(i = 0;i<n;i++)        {            scanf("%d%d",&a,&b);            while(b--)            {                val[l++] = a;//將價值存入陣列                sum+=a;            }        }        for(i = 0;i<l;i++)        {            for(j = sum/2;j>=val[i];j--)//01揹包            {                dp[j] = max(dp[j],dp[j-val[i]]+val[i]);            }        }        printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);    }    return 0;}

 

           

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