「usaco2002.4」Chores
题目描述
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.
输入
* 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).
输出
A single line with an integer which is the least amount of time required to perform all the chores.
样例输入
7 5 0 1 1 1 3 1 2 6 1 1 1 2 2 4 8 2 2 4 4 3 3 5 6
样例输出
23
提示
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[Here is one task schedule: Chore 1 starts at time 0, ends at time 5. Chore 2 starts at time 5, ends at time 6. Chore 3 starts at time 6, ends at time 9. Chore 4 starts at time 5, ends at time 11. Chore 5 starts at time 11, ends at time 12. Chore 6 starts at time 11, ends at time 19. Chore 7 starts at time 19, ends at time 23. ] |
代码
深搜水过，ed[i]，表示第i个任务的最早完成时间。。
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#include<iostream> #include<cstdio> using namespace std; int n,top,v[10001],r[10001],ed[10001],head[10001],ans,cnt; struct data{int to,next;}e[1000001]; void insert(int u,int v) { cnt++; e[cnt].to=v; e[cnt].next=head[u]; head[u]=cnt; } void dfs(int k,int t) { int i=head[k]; while(i) { if(t+v[e[i].to]>ed[e[i].to]){ed[e[i].to]=t+v[e[i].to];dfs(e[i].to,ed[e[i].to]);} i=e[i].next; } ans=max(ans,t); } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) { int x,y; scanf("%d%d",&v[i],&x); for(int j=1;j<=x;j++) { scanf("%d",&y); insert(i,y); r[y]++; } } for(int i=1;i<=n;i++) if(!r[i])insert(0,i); dfs(0,0); printf("%d",ans); return 0; } |
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