POJ1949 Chores 解题报告【拓扑排序/DAG最长路】

Description
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.
Input
* 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).
Output
A single line with an integer which is the least amount of time required to perform all the chores.
Sample Input
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
Sample Output
23
Hint
[
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.
]
解题报告
题意就是说要完成n个任务，这n个任务都有自己的先决条件，由此读入的时候需要反向建边。此外，由于这些先决条件可以同时进行，那么他们完成的时间也就是这几个之中的最大值（也就是点权的最大值）。这样一来就很明朗了。

#include<cstdio>#include<cstring>#include<algorithm>#include<queue>using namespace std;const int N=10000,M=1000000;struct edge{    int v,next;}ed[M+5];int head[N+5],num;int n,w[N+5],in[N+5],out[N+5],dis[N+5],vmax;void build(int u,int v){    ed[++num].v=v;    ed[num].next=head[u];    head[u]=num;}void toposort(){    queue<int>q;    for(int i=1;i<=n;i++)dis[i]=-0x7fffffff;    for(int u=1;u<=n;u++)    if(!in[u]){q.push(u);dis[u]=w[u];}    while(!q.empty())    {        int u=q.front();q.pop();        for(int i=head[u];i!=-1;i=ed[i].next)        {            int v=ed[i].v;            dis[v]=max(dis[v],dis[u]+w[v]);            if(!(--in[v]))q.push(v);        }    }}int main(){    memset(head,-1,sizeof(head));    scanf("%d",&n);    for(int v=1;v<=n;v++)    {        int tot;        scanf("%d%d",&w[v],&tot);        while(tot--)        {            int u;            scanf("%d",&u);            build(u,v);            in[v]++,out[u]++;        }    }    toposort();    for(int v=1;v<=n;v++)if(!out[v])vmax=max(vmax,dis[v]);    printf("%d",vmax);    return 0;}