洛谷 P3040 [USACO12JAN]贝尔分享Bale Share

题目描述

Farmer John has just received a new shipment of N (1 <= N <= 20) bales of hay, where bale i has size S_i (1 <= S_i <= 100). He wants to divide the bales between his three barns as fairly as possible.

After some careful thought, FJ decides that a "fair" division of the hay bales should make the largest share as small as possible. That is, if B_1, B_2, and B_3 are the total sizes of all the bales placed in barns 1, 2, and 3, respectively (where B_1 >= B_2 >= B_3), then FJ wants to make B_1 as small as possible.

For example, if there are 8 bales in these sizes:

2 4 5 8 9 14 15 20

A fair solution is

Barn 1: 2 9 15   B_1 = 26

Barn 2: 4 8 14   B_2 = 26

Barn 3: 5 20     B_3 = 25

Please help FJ determine the value of B_1 for a fair division of the hay bales.

FJ有N (1 <= N <= 20)包干草，干草i的重量是 S_i (1 <= S_i <= 100)，他想尽可能平均地将干草分给3个农场。

他希望分配后的干草重量最大值尽可能地小，比如， B_1,B_2和 B_3是分配后的三个值，假设B_1 >= B_2 >= B_3，则他希望B_1的值尽可能地小。

例如：8包干草的重量分别是：2 4 5 8 9 14 15 20，一种满足要求的分配方案是

农场 1: 2 9 15 B_1 = 26

农场 2: 4 8 14 B_2 = 26

农场 3: 5 20 B_3 = 25

请帮助FJ计算B_1的值。

输入输出格式

输入格式：

Line 1: The number of bales, N.
Lines 2..1+N: Line i+1 contains S_i, the size of the ith bale.

输出格式：

Line 1: Please output the value of B_1 in a fair division of the hay bales.

输入输出样例

输入样例#1：复制

输出样例#1：复制

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#define MAXN 30

using namespace std;

int n;

int A,B,C;

int ans=0x7f7f7f7fl;

int sum[MAXN];

void dfs(int now){

    if(max(A,max(B,C))>ans)    return ;

    if(now==n+){

        ans=max(A,max(C,B));

        return;

    }

    A+=sum[now];dfs(now+);A-=sum[now];

    B+=sum[now];dfs(now+);B-=sum[now];

    C+=sum[now];dfs(now+);C-=sum[now];

}

int main(){

    scanf("%d",&n);

    for(int i=;i<=n;i++)

        scanf("%d",&sum[i]);

    dfs();

    cout<<ans;

}

60分的dfs

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#define MAXN 30

using namespace std;

int n;

int A,B,C;

int ans=0x7f7f7f7fl;

int sum[MAXN];

void dfs(int now){

    if(now==n+){

        ans=max(A,max(C,B));

        return;

    }

    A+=sum[now];if(A<ans)    dfs(now+);A-=sum[now];

    B+=sum[now];if(B<ans)    dfs(now+);B-=sum[now];

    C+=sum[now];if(C<ans)    dfs(now+);C-=sum[now];

}

int main(){

    scanf("%d",&n);

    for(int i=;i<=n;i++)

        scanf("%d",&sum[i]);

    dfs();

    cout<<ans;

}

AC的dfs

正解思路：动态规划。

f[i][j][k]表示到第i堆干草为止，第一个农场分到j的干草，第二个农场分到k的干草。

#include<cstdio>

#include<cstring>

#include<iostream>

#include<algorithm>

#define MAXN 10010

using namespace std;

int n;

int dp[][20][20];

int num[MAXN],sum[MAXN];

int dfs(int now,int x,int y){

    int z=sum[now-]-x-y;

    if(now==n+)    return max(x,max(y,z));

    if(dp[now][x][y])    return dp[now][x][y];

    dp[now][x][y]=min(dfs(now+,x+num[now],y),min(dfs(now+,x,y+num[now]),dfs(now+,x,y)));

    return dp[now][x][y];

}

int main(){

    scanf("%d",&n);

    for(int i=;i<=n;i++)    scanf("%d",&num[i]);

    for(int i=;i<=n;i++)    sum[i]=sum[i-]+num[i];

    dfs(,,);

    cout<<dp[][][];

}