2017 百度之星预赛1004_百度之星2017预选赛-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_39814258/article/details/77425081

本文介绍了一个使用动态规划解决背包问题的C++程序实现。通过给定的物品价值和重量，程序能够找出不超过背包容量限制下最大价值的组合，并输出所选物品的编号。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct node
{
int money;
int value;
}num[105];
int dp[105][1005];
int vis[105];
int main()
{
int temp = 0;
int B, N, t;
scanf("%d", &t);
while (t--)
{
  scanf("%d %d", &B, &N);
  temp++;
  memset(dp, 0, sizeof(dp));
  memset(vis, 0, sizeof(vis));
  for (int i = 1; i <= N; i++)
  {
   scanf("%d %d", &num[i].value, &num[i].money);
  }
  for (int i = 1; i <= N; i++)
  {
   for (int j = B; j >= 0; j--)
   {
    if (j >= num[i].money)
    {
     dp[i][j] = max(dp[i - 1][j - num[i].money] + num[i].value, dp[i - 1][j]);
    }
    else
    {
     dp[i][j] = dp[i - 1][j];
    }
   }
  }
  int maxm = B, maxmoney = 0;
  for (int i = N; i >= 1; i--)
  {
   //cout<<dp[i][maxm]<<" "<<dp[i-1][maxm-num[i].money]<<" "<<dp[i][maxm]-num[i].value<<endl;
   if (maxm - num[i].money >= 0)
   {
    if (dp[i][maxm] - num[i].value == dp[i - 1][maxm - num[i].money] && dp[i][maxm] != dp[i - 1][maxm])
    {
     maxm -= num[i].money;
     maxmoney += num[i].money;
     vis[i] = 1;
    }
   }
  }
  printf("Case #%d:\n%d %d\n", temp, dp[N][B], maxmoney);
  if (dp[N][B])
  {
   int i = 1;
   while (!vis[i])
   {
    i++;
   }
   printf("%d", i);
   for (int j = i + 1; j <= N; j++)
   {
    if (vis[j])
     printf(" %d", j);
   }
   printf("\n");
  }
}
return 0;
}