2019ICPCshenyang网络赛(C. Dawn-K's water)

题目：

Dawn-K recently discovered a very magical phenomenon in the supermarket of Northeastern University: The large package is not necessarily more expensive than the small package.

On this day, Dawn-K came to the supermarket to buy mineral water, he found that there are nntypes of mineral water, and he already knew the price pp and the weight cc (kg) of each type of mineral water. Now Dawn-K wants to know the least money aa he needs to buy no less than mm kilograms of mineral water and the actual weight bb of mineral water he will get. Please help him to calculate them.

Input

The input consists of multiple test cases, each test case starts with a number nn (1 \le n \le 10^31≤n≤103) – the number of types, and mm (1 \le m \le 10^41≤m≤104) – the least kilograms of water he needs to buy. For each set of test cases, the sum of nn does not exceed 5e45e4.

Then followed n lines with each line two integers pp (1 \le p \le 10^91≤p≤109) – the price of this type, and cc (1 \le c \le 10^41≤c≤104) – the weight of water this type contains.

Output

For each test case, you should output one line contains the minimum cost aa and the weight of water Dawn-K will get bb. If this minimum cost corresponds different solution, output the maximum weight he can get.

(The answer aa is between 11 and 10^9109, and the answer bb is between 11 and 10^4104)

样例输入

3 3
2 1
3 1
1 1
3 5
2 3
1 2
3 3

样例输出

3 3
3 6

题意描述：这个题可以看成是一个完全背包的问题，就是要买满足至少m的水需要花费的最少钱是多少，其中n是水的种类，a[i]是水的价格，b[i]是以a[i]的价格买的水的重量。

程序代码：

#include<iostream>
#include<algorithm>
using namespace std;
long long mi=1e14;//表示样例所给的大小是10000
int a[1010],b[1010];
long long dp[50100];//给的样例是10^9，所以用long long
int main()
{
  int i,j,k,m,n,sum;
  long long x,y;
  while(scanf("%d%d",&n,&m)!=EOF)
  {
    sum=20010;
    for(i=0;i<n;i++)
      scanf("%d%d",&a[i],&b[i]);
    for(i=1;i<=sum;i++)
      dp[i]=mi;//定义最初的价格都是无穷大
    dp[0]=0;//买0需要花费0
    for(i=0;i<n;i++)
      for(j=b[i];j<=sum;j++)
      {
        dp[j]=min(dp[j],dp[j-b[i]]+a[i]);
      }
    x=dp[m];
    y=m;
    for(i=m+1;i<=sum;i++)//寻找在sum和m之间是否还有满足的其价格更低
    {
      if(dp[i]<=x)
      {
        x=dp[i];
        y=i;
      }
    } 
    printf("%lld %lld\n",x,y);  
  }
  return 0;
}