51 nod 1007 正整数分组 (简单01背包) && csu 1547: Rectangle

http://www.51nod.com/onlineJudge/questionCode.html#problemId=1007&noticeId=15020

求出n个数的和sum,然后用sum/2作为背包容量,让n个数去放,求出一个最大价值,那么这就是其中一组的和,另外一组的和就是sum-dp[sum/2];

注意这里的体积和价值都是a[i];

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1000000000
 #define N 2510
 #define mod 1000000000
 using namespace std;

 int a[],dp[];
 int main()
 {
    // freopen("a.txt","r",stdin);
     int n,sum=,n1=;
     scanf("%d",&n);
     for(int i=;i<n;i++)
     {
         scanf("%d",&a[i]);
         sum+=a[i];
     }
     n1=sum/;
     memset(dp,,sizeof(dp));
     for(int i=n-;i>=;i--)
        for(int j=n1;j>=;j--)
        {
            if(j>=a[i])
            {
                dp[j]=max(dp[j],dp[j-a[i]]+a[i]);
            }
        }
     //printf("%d\n",dp[n1]);
     printf("%d\n",abs(sum-dp[n1]-dp[n1]));
     return ;
 }

http://acm.csu.edu.cn/OnlineJudge/problem.php?id=1547

上面这题的变形,把a==1的矩形长度累加起来,然后做一次01背包,就能得知放置这些宽度为1的矩形所需要的最小长度是多少,

但是要注意可能sum/2的背包可能放不下,那就需要取两个数的最大值,最后加上宽度为2的矩形长度即可。

 #include <iostream>
 #include <cstdio>
 #include <cmath>
 #include <vector>
 #include <cstring>
 #include <string>
 #include <algorithm>
 #include <string>
 #include <set>
 #include <functional>
 #include <numeric>
 #include <sstream>
 #include <stack>
 #include <map>
 #include <queue>
 #pragma comment(linker, "/STACK:102400000,102400000")
 #define CL(arr, val)    memset(arr, val, sizeof(arr))

 #define ll long long
 #define inf 0x7f7f7f7f
 #define lc l,m,rt<<1
 #define rc m + 1,r,rt<<1|1
 #define pi acos(-1.0)

 #define L(x)    (x) << 1
 #define R(x)    (x) << 1 | 1
 #define MID(l, r)   (l + r) >> 1
 #define Min(x, y)   (x) < (y) ? (x) : (y)
 #define Max(x, y)   (x) < (y) ? (y) : (x)
 #define E(x)        (1 << (x))
 #define iabs(x)     (x) < 0 ? -(x) : (x)
 #define OUT(x)  printf("%I64d\n", x)
 #define lowbit(x)   (x)&(-x)
 #define Read()  freopen("a.txt", "r", stdin)
 #define Write() freopen("b.txt", "w", stdout);
 #define maxn 1000000000
 #define N 2510
 #define mod 1000000000
 using namespace std;
 int dp[];
 int main()
 {
     //freopen("a.txt","r",stdin);
     int t,n,a,b,c[],n1,n2;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d",&n);
         n1=n2=;
         memset(dp,,sizeof(dp));
         memset(c,,sizeof(c));
         int k=;
         for(int i=;i<n;i++)
         {
             scanf("%d%d",&a,&b);
             if(a==) {n1+=b;c[k++]=b;}
             n2+=b;
         }
        // printf("%d\n",k);
         for(int i=;i<k;i++)
             for(int j=n1/;j>=c[i];j--)
             if(j>=c[i])
             dp[j]=max(dp[j],dp[j-c[i]]+c[i]);
        // printf("%d\n",dp[n1/2]);
         printf("%d\n",max(n1-dp[n1/],dp[n1/])+n2-n1);
     }
     return ;
 }