USACO 1.3.2 Crafting Winning Solutions

Text这期特别重要= =

Notes

• Try to break the algorithm - use special (degenerate?) test cases

• 错了之后想想

  • debug多久了
  • 是否有所线索
  • 看上去错了什么类型的错误
  • A short amount (20 mins) of debugging is better than switching to anything else; but you might be able to solve another from scratch in 45 mins.
  • 什么时候回来做
  • Consider from here out - forget prior effort, focus on the future: how can you get the most points in the next hour with what you have?

• Do the Math

• 过滤,筛选,预计算,对称,

Prime Cryptarithm

暴搜一遍.

距离

一开始没看懂题意= =
环形距离

bool close(int a, int b)
{
  if (abs(a-b) <= 2) return true;
  if (abs(a-b) >= N-2) return true;
  return false;
}

wormholes

栽在这个函数上

bool cycle_exists(void)
{
    for (int start=1; start<=n; start++) {
        // does there exist a cylce starting from start
        int pos = start;
        for (int count=0; count<n; count++)
        {
            pos = net[cycle[pos]];
        }
        if (pos != 0) return true;
    }
    return false;
}

• 也就是说找循环不一定要递归去找.暴力转足够次数如果还能回来就有循环了
• 另外total的统计方式也值得琢磨
  语法trick

void test(){
    suc=0;
    for (int i=1; suc==0 && i<=n; i++) 
    ...      
    tot+=suc;
}

这样不如用bool函数= =直接return

另外解决了的坑:
- 配对:
- 配对问题不能在递归函数中双重循环 这样会重复
- 不能只复原一个cycle 这样会遗漏很多

void dfs(int k,int beg){
    if (k==n) {
       test();
    }else if(beg<n){
        if (cycle[beg]) {
            dfs(k,beg+1);
        }else {
            for (int i=1; i<=n; i++) 
                if (!cycle[i]&&i!=beg) {
                    cycle[i]=beg;
                    cycle[beg]=i;
                    dfs(k+2,beg+1);
                    cycle[i]=cycle[beg]=0;
                }
        }
    }
}

(话说youtube搜usaco居然是这道题的题解视频..不过小哥为什么不说中文呢…
b站题解

Ski Course Design

一堆数据让极差<=17,修改一个数据需要 △2 费用,问最小费用
读的时候扫最大最小,再扫一遍每个数据和最大最小的差,枚举修改最小值所需要费用.
官方题解省了两个数组

for (int j=0; j<n; j++)
        {
            // if hill is below the interval
            if (hills[j]<i)
                x=i-hills[j];
            // if hill is above the interval
            else if (hills[j]>i+17)
                x=hills[j]-(i+17);
            // if hill is int the interval
            else
                x=0;
            cost+=x*x;
        }

开始涉及算法涉及啦!加油~