POJ 2576 Tug of War 二维背包 OR 随机化

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题意:n个人n<=100,每个人有权值a[i]<=450,要求把n个人分成两组,人数差<=1&&权值差最小
dp(i)[j][k]前i个人选j个是否能凑出权值为k? n为奇数可以选half or half-1 n为偶数只能选half个复杂度O(n*n*M)

#include <iostream>#include <algorithm>#include <cstdio>#include <cstring> using namespace std;typedef long long ll;const int N=1e2+20;const int M=455;int dp[N][N*M];//dp(i)[j][k]前i个人选j个是否能凑出权值为k? int n,a[N]; int main(){while(cin>>n){int sum=0,s;memset(dp,0,sizeof(dp));dp[0][0]=1;for(int i=1;i<=n;i++)cin>>a[i],sum+=a[i];int half=(n+1)/2;s=sum;sum/=2;for(int i=1;i<=n;i++){//逆推保证dp[j-1][k-a[i]]是前i-1个人中选出的for(int j=half;j>=1;j--){for(int k=sum;k>=a[i];k--){//cout<<j<<' '<<a[i]<<' '<<k<<endl;dp[j][k]|=dp[j-1][k-a[i]];}}}int j;for(j=sum;j>=0;j--){if(n%2){if(dp[half][j]||dp[half-1][j])break;}else{if(dp[half][j])break;}}cout<<min(j,s-j)<<' '<<max(j,s-j)<<endl; }return 0;}

随机化:步骤一:随机分成两组步骤二:不断随机交换两个数若更优则交换否则继续随机直到一定次数后返回步骤1

代码参考链接

#include <stdio.h>#include <string.h>#include <stdlib.h>#include <math.h>#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))#define N 102int n,m,a[N],s[N],t[N];int main(){    int i,j,sum,s1,s2,T=250,temp1,temp2,len1,len2,flag,res1=0;//T是随机的实验组数，亲测245WA，250AC    s1 = s2 = sum = 0;    scanf("%d",&n);    for(i = 0;i<n;i++){        scanf("%d",&a[i]);        sum += a[i];    }    if(n==1){        printf("0 %d\n",a[0]);        return 0;    }    m = n - n/2;    n /= 2;    while(T--){        len1 = len2 = flag = s1 = s2 = 0;        for(i = 0;len1<n && len2<m;i++){//将数据随机分成两份            if(rand()%2){                t[len2++] = a[i];                s2 += a[i];            }            else{                s[len1++] = a[i];                s1 += a[i];            }        }        while(len1 < n){            s1 += a[i];            s[len1++] = a[i++];        }        while(len2 < m){            s2 += a[i];            t[len2++] = a[i++];        }        while(flag<10){            i = rand()%n;//在两份数据中任选一个进行调换            j = rand()%m;            temp1 = abs(s1 - s2);            temp2 = abs(s1 - s2 - 2*(s[i]-t[j]));            if(temp1 > temp2){//如果调换结果更好，那么进行调换                flag = 0;                s1 -= s[i]-t[j];                s2 -= t[j]-s[i];                temp1 = s[i];                s[i] = t[j];                t[j] = temp1;            }            flag++;//flag保存有多少次没有进行过更新了        }        res1 = max(res1,min(s1,s2));    }    printf("%d %d\n",res1,sum-res1);    return 0;}

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