poj3628-DFS/0-1背包-DP/枚舉-資料比較弱、方法比較多

因為資料範圍20，所以直接枚舉是2^20，不會逾時。直接求組合就行。在N個數裡面取1個數，2個數。。。。N個數，求出一個最小差值就可以了。

下面是組合的演算法--175MS

#include<stdio.h>#include <stdlib.h>#include <string.h>#include <math.h>#define nMax 25int N,B;int height[nMax];int ans;int getSum(int pos){int sum = 0;for (int i = 0; i < N; ++ i ){if (pos & (1 << i)){sum += height[i];}}return sum;}int main(){while (scanf("%d %d", &N, &B) != EOF){ans = 0x7FFFFFFF;for (int i = 0; i < N; ++ i){scanf("%d", &height[i]);}for (int i = 1; i < (1 << N) ; ++ i){if (getSum(i) < B){continue;}int tmp = getSum(i) - B;if (ans > tmp){ans = tmp;}}printf("%d\n", ans);}return 0;}

這道題目還可以用01背包動態規劃做--32MS

#include<iostream>   #include<cstring>   using namespace std;  int f[1000050],c[10000];  int main(){      int n,v;      while(cin>>n>>v){          memset(f,0,sizeof(f));          memset(c,0,sizeof(c));          int sum=0;          for(int i=1;i<=n;++i){              cin>>c[i];              sum+=c[i];          }          for(int i=1;i<=n;i++){              for(int j=sum;j>=c[i];j--)                  f[j]=max(f[j],f[j-c[i]]+c[i]);          }          for(int i=1;i<=sum;++i){              if(f[i]>=v){                  cout<<f[i]-v<<endl;                  break;              }          }      }      return 0;     }  

 

這道題目還可以用dfs深度優先搜尋做--32MS

#include<stdio.h>   int cow[20];  int b;  int n;  int ans=99999999;  void DFS(int num,int sum)  {      if(sum>=ans) return;      if(num==n)      {          if(sum>=b) ans=sum;          return;      }      DFS(num+1,sum);      DFS(num+1,sum+cow[num]);  }  int main()  {      //freopen("input","r",stdin);       int i,j,cnt,tmp;      int sum;      scanf("%d %d",&n,&b);      for(i=0;i<n;i++)          scanf("%d",&cow[i]);      DFS(0,0);      printf("%d",ans-b);      return 0;  }