Weights and Measures - UVa 10154 dp

Problem F: Weights and Measures

I know, up on top you are seeing great sights, 
But down at the bottom, we, too, should have rights. 
We turtles can't stand it. Our shells will all crack! 
Besides, we need food. We are starving!" groaned Mack.

The Problem

Mack, in an effort to avoid being cracked, has enlisted your advice as to the order in which turtles should be dispatched to form Yertle's throne. Each of the five thousand, six hundred and seven turtles ordered by Yertle has a different weight and strength. Your task is to build the largest stack of turtles possible.

Standard input consists of several lines, each containing a pair of integers separated by one or more space characters, specifying the weight and strength of a turtle. The weight of the turtle is in grams. The strength, also in grams, is the turtle's overall carrying capacity, including its own weight. That is, a turtle weighing 300g with a strength of 1000g could carry 700g of turtles on its back. There are at most 5,607 turtles.

Your output is a single integer indicating the maximum number of turtles that can be stacked without exceeding the strength of any one.

Sample Input

300 1000
1000 1200
200 600
100 101

Sample Output

3

题意：每只乌龟有重量和承受力，每只乌龟上面的重量加上自身的重量不能超过他的承受力，问最多能叠多少只乌龟。

思路：每只乌龟先按承受力排序从小到大，承受力相同时按重量从小到大排序，然后dp[i]表示叠i只乌龟时的最小重量。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
struct node
{ ll w,s;
}box[6010];
ll dp[6010];
bool cmp(node a,node b)
{ if(a.s==b.s)
   return a.w<b.w;
  return a.s<b.s;
}
int main()
{ int n=1,i,j,k,len=1;
  while(~scanf("%lld%lld",&box[n].w,&box[n].s))
  { if(box[n].s<box[n].w)
     continue;
    n++;
  }
  n--;
  sort(box+1,box+1+n,cmp);
  dp[1]=box[1].w;
  for(i=2;i<=n;i++)
  { if(dp[len]<=box[i].s-box[i].w)
     dp[++len]=dp[len-1]+box[i].w;
    for(j=len-1;j>=0;j--)
     if(dp[j]<=box[i].s-box[i].w)
      dp[j+1]=min(dp[j+1],dp[j]+box[i].w);
  }
  if(n==0)
   printf("0\n");
  else
  printf("%d\n",len);
}