MUTC 2 E - Save the dwarfs DP?

Save the dwarfs

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 934    Accepted Submission(s): 285

Problem Description

Several dwarfs are trapped in a deep well. They are not tall enough to climb out of the well, so they want to make a human-pyramid, that is, one dwarf stands on another's shoulder, until the dwarf on the top can reach the top of the well when he raise his arms up. More precisely speaking, we know the i-th dwarf's height from feet to shoulder is Ai, and his arm length Bi. And we know the height of the well is H. If we can build a dwarf-tower consists of dwarf 1, dwarf 2, ..., dwarf k from bottom to top, such that A1 + A2 + ... + Ak-1 + Ak + Bk >= H, then dwarf k can escape from the well. Obviously, the escaped dwarf can't be used to build tower again.

We want the escaped dwarfs as many as possible. Please write a program to help the dwarfs.

 

Input

The first line of each test case contains N, (1 <= N <= 2000) the number of trapped dwarfs. Each of the following N lines contains two integers Ai and Bi. The last line is H, the height of the well. All the integers are less than 100,000.

 

Output

Output one line for each test case, indicating the number of dwarfs escaped at most.

 

Sample Input

2
20 10
5 5
30
2
20 10
5 5
35

 

Sample Output

2
1
Hint
For the first case, the tall dwarf can help the other dwarf escape at first. For the second case, only the tall dwarf can escape.
 

 

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DP?

转自标准题解

dp[i][j]表示最后i个人能逃出j个时，需要之前井中剩下的人的最小A高度之和。

d[]按a+b从大到小排序

dp[i][j] = min(dp[i-1][j] - d[i].a, max(dp[i-1][j-1], h - sumA[i] -s[i].b  ))

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#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn=2222;
const int INF=1e9;

int n,h;
struct dwarfs{
    int a;
    int b;
}d[maxn];
int f[maxn][maxn];
int sum[maxn];
int ans;
bool cmp(dwarfs x,dwarfs y)
{
    return (x.a+x.b)>(y.a+y.b);
}

int main()
{
    while (~scanf("%d",&n))
    {
        memset(f,0,sizeof(f));
        sum[0]=0;
        for (int i=1;i<=n;i++)
        {
            scanf("%d%d",&d[i].a,&d[i].b);
        }
        scanf("%d",&h);
        sort(d+1,d+n+1,cmp);
        for (int i=1;i<=n;i++)
        {
            sum[i]=sum[i-1]+d[i].a;
        }
        for (int i=0;i<=n;i++)
        {
            for (int j=0;j<=n;j++)
            {
                f[i][j]=INF;
            }
        }
        f[0][0]=0;
        for (int i=1;i<=n;i++)
        {
            for (int j=0;j<=i;j++)
            {
                f[i][j]=f[i-1][j]-d[i].a;
                int tmp;
                if (j>0) tmp=max(f[i-1][j-1],h-sum[i]-d[i].b);
                else tmp=h-sum[i]-d[i].b;
                if (tmp<f[i][j]) f[i][j]=tmp;
                if (f[i][j]<0) f[i][j]=0;
            }
        }
        for (int j=n;j>=0;j--)
        {
            if (f[n][j]<=0)
            {
                ans=j;
                break;
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}