51Nod 1007 正整数分组 01背包

来源：互联网发布：淘宝香辣零食店铺推荐编辑：程序博客网时间：2024/06/06 18:42

51Nod 1007 正整数分组 01背包

题目链接：51Nod 1007 正整数分组。
思路：假定2组分别为集合A，集合B，并且集合A的和SUM(A)小于或等于集合B的和SUM(B)。对于每个数组，只有在A集合和不在A集合(或者是在B集合)的情况。然后可以将题目转化为01背包模型：在[SUM(A)+SUM(B)]/2【PS:这里是整数除法】的容量下，所能取得的最大价值，每个物品的价值和体积都是a[i]。

#include <cmath>#include <queue>#include <vector>#include <cstdio>#include <string>#include <cstring>#include <iomanip>#include <iostream>#include <algorithm>using namespace std;//#pragma comment(linker, "/STACK:1024000000,1024000000")#define FIN             freopen("input.txt","r",stdin)#define FOUT            freopen("output.txt","w",stdout)#define fst             first#define snd             second#define lson            l, mid, rt << 1#define rson            mid + 1, r, rt << 1 | 1// typedef __int64  LL;typedef long long LL;typedef unsigned int uint;const int INF = 0x3f3f3f3f;const double eps = 1e-6;const int MAXN = 1000000 + 5;const int MAXM = 10000 + 5;int N;int A[MAXN];int dp[MAXN];int main() {#ifndef ONLINE_JUDGE    FIN;#endif // ONLINE_JUDGE    while (~scanf("%d", &N)) {        int SUM = 0;        for (int i = 0; i < N; i ++) {            scanf("%d", &A[i]);            SUM += A[i];        }        memset(dp, 0, sizeof(dp));        int W = SUM >> 1;        for (int i = 0; i < N; i ++) {            for (int j = W; j >= A[i]; j --) {                dp[j] = max(dp[j], dp[j - A[i]] + A[i]);            }        }        int a = dp[W], b = SUM - a;        printf("%d\n", b - a);    }    return 0;}

0 0