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. 

#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！

#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！

#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！