Codeforces Round #178 (Div. 2) B. Shaass and Bookshelf —— DP

题目链接：http://codeforces.com/contest/294/problem/B

B. Shaass and Bookshelf

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Shaass has n books. He wants to make a bookshelf for all his books. He wants the bookshelf's dimensions to be as small as possible.
The thickness of the i-th book is ti and
its pages' width is equal to wi.
The thickness of each book is either 1 or 2.
All books have the same page heights.

Shaass puts the books on the bookshelf in the following way. First he selects some of the books and put them vertically. Then he puts the rest of the books horizontally above the vertical books. The sum of the widths of the horizontal books must be no more
than the total thickness of the vertical books. A sample arrangement of the books is depicted in the figure.

Help Shaass to find the minimum total thickness of the vertical books that we can achieve.

Input

The first line of the input contains an integer n, (1 ≤ n ≤ 100).
Each of the next n lines contains two integers ti and wi denoting
the thickness and width of the i-th book correspondingly, (1 ≤ ti ≤ 2, 1 ≤ wi ≤ 100).

Output

On the only line of the output print the minimum total thickness of the vertical books that we can achieve.

Examples

input

5
1 12
1 3
2 15
2 5
2 1

output

5

input

3
1 10
2 1
2 4

output

3

题解：

数据范围很小，所以可以直接开三维数组。

1.dp[i][j][k]表示：第i本书，下面为j， 上面为k的状态是否存在。

2.对于每一个已经存在的状态，再去判断当前的书是否可以放上去。

代码如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 100+10;

int n;
int w[maxn], t[maxn];
int dp[maxn][maxn<<1][maxn<<1];

int main()
{
    scanf("%d",&n);
    int sum = 0;
    for(int i = 1; i<=n; i++)
        scanf("%d%d", &t[i], &w[i]), sum += t[i];

    dp[0][sum][0] = 1;  //初始化全部放在下面
    for(int i = 1; i<=n; i++)
    for(int j = sum; j>=0; j--)
    for(int k = 0; k<=j; k++)
    {
        if(!dp[i-1][j][k]) continue;  //如果放在下面的状态不存在，则直接退出

        dp[i][j][k] = 1;    //留在下面
        if(j-t[i]>=k+w[i])  //放上去
            dp[i][j-t[i]][k+w[i]] = 1;
    }

    int ans = INF;
    for(int j = sum; j>=0; j--)
    for(int k = 0; k<=j; k++)
        if(dp[n][j][k])
            ans = min(ans,j);
    cout<< ans <<endl;
    return 0;
}

或者用记忆化搜索写：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const double eps = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int maxn = 100+10;

int n;
int w[maxn], t[maxn];
int dp[maxn][maxn<<1][maxn<<1];

int dfs(int pos, int thi, int wid)
{
    if(pos==n+1) return thi;
    if(dp[pos][thi][wid]!=-1) return dp[pos][thi][wid];

    if(thi-t[pos]>=w[pos]+wid)
        return dp[pos][thi][wid] = min( dfs(pos+1, thi-t[pos], w[pos]+wid), dfs(pos+1, thi, wid) );
    else
        return dp[pos][thi][wid] = dfs(pos+1, thi, wid);
}

int main()
{
    scanf("%d",&n);
    int sum = 0;
    for(int i = 1; i<=n; i++)
        scanf("%d%d", &t[i], &w[i]), sum += t[i];

    memset(dp,-1,sizeof(dp));
    cout<< dfs(1, sum, 0) <<endl;
    return 0;
}