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 }