Jobdu 1420:Jobdu MM分水果 - KingEasternSun的专栏 - 博客频道 - CSDN.NET


九度笔记之 1420:Jobdu MM分水果 - KingEasternSun的专栏 - 博客频道 - CSDN.NET
Jobdu团队有俩PPMM,这俩MM干啥都想一样。一天,富强公司给团队赞助了一批水果,胡老板就把水果派发给了这俩MM,由她们自行分配。每个水果都有一个重量,你能告诉她们怎么分才使得分得的重量差值最小吗?
     我们利用动态规划来做。假设所有水果总重为sum, 那么其中一个人所能分得的水果重量w范围一定在 0 - sum/2之间,
     也许有人会问 sum为奇数时,w范围应该在0 - (sum/2+1)之间,当一个人分得sum/2+1时,另外一个人分得的必定是sum/2,这和我们求质数时,测试范围从 2到sqrt(n)一样。
     我们的问题就转化为  一个人所分得的重量w从 sum/2 到0  哪个重量可以实现。
     我们当给一个人什么水果都不分的话,就可以实现0.  即 dp[0] = true;
     当我们选择第i个水果时,该水果重量为num[i], 我们判断在第i个水果分配给这个人之前 dp[w]是否为true,或者dp[w-num[j]]是否为true;
            如果从前i-1个水果中挑出某几个可以组成重量w,那么dp[w] = true;
            如果从前i-1个水果中挑出某几个可以组成重量w-num[i], 那么加上第i个水果的重量就可以组成w,所以
  1. dp[j] = dp[j]||dp[j-num[i]];  
      在更新dp的时候一定注意是要从sum/2 到 num[i], 这样才能保证每个水果出现了0次或1次。

  1. void allocate(int n){  
  2.     int sum=0;  
  3.     for(int i = 0;i<n;i++){  
  4.         std::cin>>num[i];  
  5.         sum+=num[i];  
  6.     }  
  7.        
  8.     int maxPercent = sum>>1;  
  9.     bool *dp = new bool[maxPercent+1];  
  10.     for(int i = 1;i<maxPercent+1;i++)  
  11.         dp[i] = false;  
  12.     dp[0] = true;  
  13.     for(int i = 0;i<n; i++){  
  14.         for(int j = maxPercent; j>=num[i]; j--)  
  15.             dp[j] = dp[j]||dp[j-num[i]];  
  16.     }  
  17.     for(int i = maxPercent;i>0;i--){  
  18.         if(dp[i]){  
  19.             std::cout<< sum-2*i<<std::endl;  
  20.             break;  
  21.         }  
  22.     }  
  23. }  
X. Brute Force
03int n, k, a[101], v[101], t,i,m;
04void dfs(int x, int y) {
05    int i;
06    if (y > t)
07        return;
08    if (y > k) {
09        k = y;
10    }
11    if (k == t)
12        return;
13    for (i = x; i <= n; i++)
14        if (!v[i]) {
15            v[i] = 1;
16            dfs(i + 1, y + a[i]);
17            v[i] = 0;
18        }
19}
20int main() {
21    while(scanf("%d",&n)==1) {
22        t=0;
23        for(i=1;i<=n;i++) {
24            scanf("%d",a+i);t+=a[i];
25        }
26        k=-1;m=t;t/=2;memset(v,0,sizeof(v));
27        dfs(1,0);
28        printf("%d\n",m-k*2);
29    }}
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