SSL 2290 - 潜水员


https://www.cnblogs.com/z360/p/6365994.html
潜水员为了潜水要使用特殊的装备。他有一个带2种气体的气缸:一个为氧气,一个为氮气。让潜水员下潜的深度需要各种的数量的氧和氮。潜水员有一定数量的气缸。每个气缸都有重量和气体容量。潜水员为了完成他的工作需要特定数量的氧和氮。他完成工作所需气缸的总重的最低限度的是多少?
    例如:潜水员有5个气缸。每行三个数字为:氧,氮的(升)量和气缸的重量:
    3 36 120
    10 25 129
    5 50 250
    1 45 130
    4 20 119
    如果潜水员需要5升的氧和60升的氮则总重最小为24912或者45号气缸)。
    你的任务就是计算潜水员为了完成他的工作需要的气缸的重量的最低值。
【输入格式】
  第一行有2整数m,n1<=m<=21,1<=n<=79)。它们表示氧,氮各自需要的量。
  第二行为整数k1<=n<=1000)表示气缸的个数。
  此后的k行,每行包括aibici1<=ai<=211<=bi<=791<=ci<=8003整数。这些各自是:第i个气缸里的氧和氮的容量及汽缸重量。
【输出格式】
  仅一行包含一个整数,为潜水员完成工作所需的气缸的重量总和的最低值


scanf("%d%d",&m,&n);

scanf("%d",&k);

for(int i=1;i<=k;i++)

   scanf("%d%d%d",&a[i],&b[i],&c[i]);

memset(f,127,sizeof(f));

f[0][0]=0;

for(int i=1;i<=k;i++)

  for(int j=m;j>=0;j--)//枚举氧含量

     for(int k=n;k>=0;k--)//枚举氮含量

        {

         int t1=j+a[i],t2=k+b[i];//含量

         if(t1>m) t1=m;//若含量超过需求,直接用需求代替

         if(t2>n) t2=n;

         if(f[t1][t2]>f[j][k]+c[i])  

           f[t1][t2]=f[j][k]+c[i];

}

https://blog.csdn.net/drtlstf/article/details/83588021
dp[0][0] = 0;
// dp[i][j] 表示还需i份氮气和j份氧气的最小重量
for (i = 1; i <= n; ++i)
{
scanf("%d%d%d", &a, &b, &c);
for (j = t; ~j; --j)
{
x = min(j + a, t);
for (k = t2; ~k; --k)
{
y = min(k + b, t2); // 确保不会超出范围
dp[x][y] = min(dp[x][y], dp[j][k] + c);
}
}

https://blog.csdn.net/qq_39670434/article/details/79476630
这道题目是不一样的二维费用的背包。或者更形象的说是二维收益,一维费用,我们要做的是达到二位收益,但是最小化一维费用。 
int f[maxn][maxn];
void DP()
{
    memset(f,10,sizeof f);
    f[0][0]=0;
    for(int i=1;i<=n;i++)
    //枚举物品
        for(int j=T;j>=0;j--)
            for(int k=A;k>=0;k--)
            //01背包从后往前枚举容量
            {
                int t1=j+t[i];
                int a1=k+a[i];
                if(t1>T)t1=T;
                if(a1>A)a1=A;
                //这里的注意点
                //如果超过了就直接赋值最大值。
                //因为收益不能多不能少
                f[t1][a1]=min(f[t1][a1],f[j][k]+w[i]);
            }
    printf("%d\n",f[T][A]);
}




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