ZZULI 1788: 小金刚的宝藏

来源：互联网发布：windows多线程同步编辑：程序博客网时间：2024/04/29 03:24

题目地址：点击打开链接

思路：价值和的一半为一个背包，往里面放东西就行了，刚开始输出搞错了，结果用搜索做超时了

AC代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>typedef long long ll;using namespace std;int value[110];int dp[100010];int main(){    int t,n;    int i,j,sum;    scanf("%d",&t);    while(t--)    {        memset(dp,0,sizeof(dp));        sum = 0;        scanf("%d",&n);        for(i=0; i<n; i++)        {            scanf("%d",&value[i]);            sum += value[i];        }        int m = sum / 2;        for(i=0; i<n; i++)        {            for(j=m; j>=value[i]; j--)            {                dp[j] = max(dp[j],dp[j-value[i]] + value[i]);            }        }        printf("%d\n",abs(sum-2*dp[m]));    }    return 0;}

超时代码：

#include <iostream>#include <cstdio>#include <cstdlib>#include <algorithm>#include <queue>#include <stack>#include <map>#include <cstring>#include <climits>#include <cmath>#include <cctype>typedef long long ll;using namespace std;int m,n,sum;int value[110];int visit[110];int flag;void dfs(int st,int sum1){    int i;    for(i=st; i<n; i++)    {        if(!visit[i])        {            if(sum1 + value[i] < m)            {                visit[i] = 1;                dfs(i+1,sum1+value[i]);                visit[i] = 0;            }            if(sum1 + value[i] >= m)            {                if(abs(sum-2*(sum1+value[i])) < flag)                    flag = abs(sum-2*(sum1+value[i]));                while(value[i] == value[i+1])                    i++;            }        }    }}int main(){    int t;    int i;    scanf("%d",&t);    while(t--)    {        memset(visit,0,sizeof(visit));        sum = 0;        scanf("%d",&n);        for(i=0; i<n; i++)        {            scanf("%d",&value[i]);            sum += value[i];        }        m = sum / 2;        flag = 10000000;        dfs(0,0);        printf("%d\n",flag);    }    return 0;}

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