poj 1949 Chores (拓扑排序+dp)

http://poj.org/problem?id=1949

题意:

有n个任务,第i个任务需要时间xi来完成,并且第i个任务必须在它 “前面的” 某些任务完成之后才能开始。

给你任务信息,问你最短需要多少时间来完成任务。

思路:

拓扑排序+dp

拓扑排序的过程中做个dp,可以保证dp的顺序...

对于这道题,找出每条依赖链,然后做dp即可。

/* ***********************************************
Author :111qqz
Created Time :2017年11月07日 星期二 13时52分27秒
File Name :3249.cpp
************************************************ */
 1#include <iostream>
 2#include <cmath>
 3#include <queue>
 4#include <cstdio>
 5#include <cstring>
 6#define PB push_back
 7#define fst first
 8#define sec second
 9#define lson l,m,rt<<1
10#define rson m+1,r,rt<<1|1
11#define ms(a,x) memset(a,x,sizeof(a))
12typedef long long LL;
13#define pi pair < int ,int >
14#define MP make_pair
 1using namespace std;
 2const double eps = 1E-8;
 3const int dx4[4]={1,0,0,-1};
 4const int dy4[4]={0,-1,1,0};
 5const int inf = 0x3f3f3f3f;
 6const int N=1E5+7;
 7vector <int>edge[N];
 8int in[N],out[N];
 9int n;
10int dp[N];
11int val[N];
12void init()
13{
14    for ( int i = 0 ; i < N ; i++) edge[i].clear();
15    ms(in,0);
16    ms(out,0);
17    ms(dp,0xca);
18 //   cout<<"dp:"<<dp[1]<<endl;
19}
20void topo()
21{
22    queue<int>Q;
23    for ( int i = 1 ; i <= n ; i++)
24    if (in[i]==0) Q.push(i),dp[i] = val[i];
 1    while (!Q.empty())
 2    {
 3    int cur = Q.front();
 4//  cout<<"cur:"<<cur<<endl;
 5    Q.pop();
 6    int siz = edge[cur].size();
 7    for ( int i = 0 ; i < siz ; i++)
 8    {
 9        int v = edge[cur][i];
10//      printf("%d->%d val[v]:%d dp[cur]: %d dp[v]:%d ",cur,v,val[v],dp[cur],dp[v]);
11        dp[v] = max(dp[v],dp[cur]+val[v]);
12//      printf("new dp[v]:%d\n",dp[v]);
13        in[v]--;
14        if (in[v]==0) Q.push(v);
15    }
16    }
17}
18void pr()
19{
20    for ( int i = 1 ; i <= n ; i++) printf("%d%c",dp[i],i==n?'\n':' ');
21}
22int main()
23{
24#ifndef  ONLINE_JUDGE 
25    freopen("./in.txt","r",stdin);
26#endif
27    while (~scanf("%d",&n))
28    {
29    init();
30    for ( int i = 1 ; i <= n ; i++)
31    {
32        int x,m;
33        x = i;
34        scanf("%d %d",&val[i],&m);
35        while (m--)
36        {
37        int y;
38        scanf("%d",&y);
39        //      cout<<"x:"<<x<<" y:"<<y<<endl;
40        edge[y].PB(x);
41        out[y]++;
42        in[x]++;
43        }
44    }
45    topo();
46    int ans = -inf;
47//  pr();
48    for ( int i = 1 ; i <= n ; i++) if (!out[i]&&dp[i]>ans) ans = dp[i];
49    printf("%d\n",ans);
50    }
1#ifndef ONLINE_JUDGE  
2    fclose(stdin);
3#endif
4    return 0;
5    }