usaco 2002 月赛 Chores 题解

Farmer John's family pitches in with the chores during milking, doing all the chores as quickly as possible. At FJ's house, some chores cannot be started until others have been completed, e.g., it is impossible to wash the cows until they are in the stalls. 

Farmer John has a list of N (3 <= N <= 10,000) chores that must be completed. Each chore requires an integer time (1 <= length of time <= 100) to complete and there may be other chores that must be completed before this chore is started. We will call these prerequisite chores. At least one chore has no prerequisite: the very first one, number 1. Farmer John's list of chores is nicely ordered, and chore K (K > 1) can have only chores 1,.K-1 as prerequisites. Write a program that reads a list of chores from 1 to N with associated times and all perquisite chores. Now calculate the shortest time it will take to complete all N chores. Of course, chores that do not depend on each other can be performed simultaneously.

* Line 1: One integer, N 

* Lines 2..N+1: N lines, each with several space-separated integers. Line 2 contains chore 1; line 3 contains chore 2, and so on. Each line contains the length of time to complete the chore, the number of the prerequisites, Pi, (0 <= Pi <= 100), and the Pi prerequisites (range 1..N, of course). 

A single line with an integer which is the least amount of time required to perform all the chores. 

       原题很简单，就是给定N个任务的时间和每个任务的前驱，求最短完成时间。

       初步研究，此题可以用最长路或是拓扑图来求解。

        vijos上有一道“休息中的小呆”，原理和此类似，然而那儿N<=100，N^3也绰绰有余，我就用floyd算法求出最长路并记录。然而现在的N<=10000，就连N^2也很危险。

        第一次，我打算也是用最长路求解。咨询过鼎神，目前只有SPFA可以有效处理最长路问题。方法就是把每条边的权值取相反数并做一遍最短路，答案再取相反数。因为N有10000而边数不知道，只好开一个边表记录。

代码一（SPFA最长路+边表优化）

#include<stdio.h>
#include<cstring>
using namespace std;
const long maxn=10001;const long INF=1;
bool flag[maxn];long cnt,i,n,j,xx,time,y,h,t,go,now,ans,tong;
long dis[maxn],begin[maxn],end[maxn],x[200*maxn];
struct arr{long l,r,s,next;}a[200*maxn];
void make_up(long l,long r,long v)
{
  a[++cnt].l=l;a[cnt].r=r;a[cnt].s=-v;a[cnt].next=-1;
  if (begin[l]==0) {begin[l]=cnt;end[l]=cnt;}
  else {a[end[l]].next=cnt;end[l]=cnt;}
}
int main()
{
  //freopen("chores.in","r",stdin);freopen("chores.out","w",stdout);
  scanf("%ld",&n);
  for (i=1;i<=n;i++)
  {
    scanf("%ld",&time);
    scanf("%ld",&xx);
    for (j=1;j<=xx;j++)
    {
      scanf("%ld",&y);
      make_up(y,i,time);
    }
    if (xx==0) make_up(0,i,time);
  }
  memset(flag,0,sizeof(flag));memset(dis,INF,sizeof(dis));
  h=0;t=1;x[1]=0;dis[0]=0;flag[0]=true;
  while (h<t)
  {
     now=x[++h];if (begin[now]==0) continue;i=begin[now];
     while (true)
     {
       go=a[i].r;
       if (dis[now]+a[i].s<dis[go])
       {
         dis[go]=dis[now]+a[i].s;
         if (!flag[go])
         {
           flag[go]=true;
           x[++t]=go;
         }
       }
       if (a[i].next==-1) break;i=a[i].next;
     }
     flag[now]=false;
  }
  for (i=1;i<=n;i++)
    if (-dis[i]>ans) ans=-dis[i];
  
  printf("%ld",ans);
  //scanf("%ld",&n);
  return 0;
}

        然而交了之后一直TLE，自己下了个数据，发现最后一个点大概要13s左右！想不到在稠密图里，SPFA的效率又如此之低！（边表的常数又很大）无论怎么优化都不行！

       

       最后只能请教互联网了，没想到，其他大牛的代码如此简单！他们用了近似DP的算法！

代码二（据网上思路改编代码）

#include<stdio.h>
#include<cstring>
using namespace std;
long f[10001],n,i,j,max,ans,xx,y;
int main()
{
  //freopen("chores.in","r",stdin);freopen("chores.out","w",stdout);
  scanf("%ld",&n);
  for (i=1;i<=n;i++)
  {
    scanf("%ld",&f[i]);
    scanf("%ld",&xx);max=0;
    for (j=1;j<=xx;j++)
    {
      scanf("%ld",&y);
      if (f[y]>max) max=f[y];
    }
    f[i]+=max;
    if (f[i]>ans) ans=f[i];
  }
  printf("%ld",ans);
  //scanf("%ld",&n);
  return 0;
}

      然而仔细一想，我发现他们只是钻了一个数据的漏洞——刚好数据的前后关系是由小到大的。思考了很长时间，我研究出了一个更加高级的算法——记忆化深搜+边表优化！

代码三（最终AC的代码）

#include<stdio.h>
#include<cstring>
using namespace std;
const long maxn=10001;const long INF=1;
long time[maxn],dp[maxn],begin[maxn],end[maxn],cnt,j,n,i,x,y,ans;
struct arr{long l,r,next;}a[200*maxn];
void make_up(long l,long r)
{
  a[++cnt].l=l;a[cnt].r=r;a[cnt].next=-1;
  if (begin[l]==0) {begin[l]=cnt;end[l]=cnt;}
  else {a[end[l]].next=cnt;end[l]=cnt;}
}
long go(long k)
{
  if (dp[k]>0) return dp[k];
  long now=begin[k];
  while (now>0)
  {
    long temp=go(a[now].r);
    dp[k]=(temp>dp[k])?temp:dp[k];
    now=a[now].next;
  }
  dp[k]+=time[k];
  return dp[k];
}
int main()
{
  //freopen("chores.in","r",stdin);freopen("chores.out","w",stdout);
  scanf("%ld",&n);
  for (i=1;i<=n;i++)
  {
    scanf("%ld",&time[i]);
    scanf("%ld",&x);
    for (j=1;j<=x;j++)
    {
      scanf("%ld",&y);make_up(i,y);
    }
    if (x==0) dp[i]=time[i];
  }
  for (i=1;i<=n;i++)
  {
    long temp=go(i);
    ans=(temp>ans)?temp:ans;
  }
  printf("%ld",ans);
  //scanf("%ld",&n);
  return 0;
}

希望众神牛看到后能够留言指导！