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

时间：2023-03-09 13:32:21
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 ;

 }

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