CodeForces990F Flow Control

Posted 2020-11-21 fang-hao

题目链接

Luogu & CodeForces

 

题目描述

You have to handle a very complex water distribution system. The system consists of $n$ junctions and $m$ pipes, $i$ -th pipe connects junctions $x_i$ and $y_i$ .

The only thing you can do is adjusting the pipes. You have to choose $m$ integer numbers $f_1$ , $f_2$ , ..., $f_m$ and use them as pipe settings. $i$ -th pipe will distribute $f_i$ units of water per second from junction $x_i$ to junction $y_i$ (if $f_i$ is negative, then the pipe will distribute $|f_i|$ units of water per second from junction $y_i$ to junction $x_i$ ). It is allowed to set $f_i$ to any integer from $-2\cdot {10}^9$ to $2\cdot {10}^9$ .

In order for the system to work properly, there are some constraints: for every $i\in [1,n]$ , $i$ -th junction has a number $s_i$ associated with it meaning that the difference between incoming and outcoming flow for $i$ -th junction must be exactly $s_i$ (if $s_i$ is not negative, then $i$ -th junction must receive $s_i$ units of water per second; if it is negative, then $i$ -th junction must transfer $|s_i|$ units of water per second to other junctions).

Can you choose the integers $f_1$ , $f_2$ , ..., $f_m$ in such a way that all requirements on incoming and outcoming flows are satisfied?

 

题目翻译

现在有一幅图，节点由水管连接起来，现在每个水管里都有一些水流，请问能否使每个节点的输入或输出的水流大小等于指定的大小。

 

解题思路

首先，肯定得判断能不能构造。

那么我们怎么判断呢？

假设一条水管连通x和y两个点，那么假设这条水管内的水流量为a，从x到y，那么x的水流量应该-a，而y的水流量应该+a，显然，两个节点的水量的和应该使0，那么我们直接判断每个节点的水量之和是不是0就好了。

接下来就是构造了，每两个节点之间都可能有很多条路径能够到达，我们只取一条，也就是构造一棵树，然后从最底层向上推，推出每根水管的水量。

 

代码

 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 #include<vector>
 6 #include<map>
 7 using namespace std;
 8 const int maxn=400050;
 9 int n,m,cnt=0;
10 struct nod{
11     int t,val;
12 };
13 vector<nod>v[maxn];
14 int jun[maxn],ans[maxn];
15 bool vis[maxn];
16 map<int,int>ma[400050];
17 inline void read(register int &x){
18     int f=1;
19     x=0; register char ch=getchar();
20     while(ch<‘0‘||ch>‘9‘){
21         if(ch==‘-‘)f=-1;
22         ch=getchar();
23     }
24     while(ch>=‘0‘&&ch<=‘9‘)x=x*10+ch-‘0‘,ch=getchar();
25     x=x*f;
26 }
27 inline void dfs(int now,int fa,int num){
28     vis[now]=1;
29     int tot=0,flag=0;
30     for(register int i=0;i<v[now].size();i++){
31         if(v[now][i].t==fa||vis[v[now][i].t])continue;
32         flag=1;
33         dfs(v[now][i].t,now,i);
34         tot+=v[now][i].val;
35     }
36     if(!flag){
37         ans[ma[fa][now]]=jun[now];
38         v[fa][num].val=jun[now];
39         return;
40     }
41     ans[ma[fa][now]]=jun[now]+tot;
42     v[fa][num].val=jun[now]+tot;
43 }
44 int main(){
45     read(n);
46     long long tot=0;
47     for(register int i=1;i<=n;i++){
48         read(jun[i]);
49         tot+=jun[i];
50     }
51     read(m);
52     register int f,t;
53     for(register int i=1;i<=m;i++){
54         read(f),read(t);
55         ma[f][t]=i;
56         ma[t][f]=i;
57         nod temp;
58         temp.t=t,temp.val=0;
59         v[f].push_back(temp);
60         temp.t=f;
61         v[t].push_back(temp);
62     }
63     if(tot!=0){
64         cout<<"Impossible"<<endl;
65         return 0;
66     }
67     nod temp;
68     temp.t=1;
69     v[0].push_back(temp);
70     dfs(1,0,0);
71     cout<<"Possible"<<endl;
72     for(register int i=1;i<=n;i++){
73         printf("%d
",ans[i]);
74     }
75 }