POJ1949 (DAG图的最长路径）_dag 图最长路题目-优快云博客

本文介绍了一种求解在给定家务任务及其先决条件的情况下，如何计算完成所有任务所需的最短时间的算法。通过使用广度优先搜索（BFS），文章详细展示了如何构建任务依赖关系图，并有效地计算出最优解。

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Chores

Time Limit: 3000MS Memory Limit: 30000K
Total Submissions: 5834 Accepted: 2758

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
解法很多。这里只是想学习一下最长路径
代码：

#include<iostream>
#include <stdio.h>
#include <stdlib.h>
#include <queue>
#include <vector>
#include <string.h>
#include <algorithm>
using namespace std;
const int N=10005;
vector<int>ch[N];
int c[N];
int dp[N];
bool vis[N];
queue<int>q;
int x,y,n,k;
int bfs()
{
    memset(vis,false,sizeof(vis));
    memset(dp,0,sizeof(dp));
    int ans=0;
    q.push(0);
    while(!q.empty())
    {
        int x=q.front();q.pop();
        vis[x]=false;
        int up=ch[x].size();
        for(int i=0;i<up;i++)
        {
            int y=ch[x][i];
            if(dp[y]>=dp[x]+c[y]) continue;
            dp[y]=dp[x]+c[y];
            if(ch[y].size()==0)
                ans=max(ans,dp[y]);
            else if(!vis[y])
            {
                q.push(y);
                vis[y]=true;
            }
        }
    }
    return ans;
}
int main()
{
    while(~scanf("%d",&n))
    {
        for(int i=0;i<=n;i++)
            ch[i].clear();
            for(int i=1;i<=n;i++)
            {
                scanf("%d%d",&c[i],&k);
                if(k==0)
                    ch[0].push_back(i);
                else
                {
                    while(k--)
                    {
                        scanf("%d",&x);
                        ch[x].push_back(i);
                    }
                }
            }
            printf("%d\n",bfs());
    }
}