题目链接
题目描述
You have to handle a very complex water distribution system. The system consists of nn
junctions and mm
pipes, ii
-th pipe connects junctions xixi
and yiyi
.
The only thing you can do is adjusting the pipes. You have to choose mm
integer numbers f1f1
, f2f2
, ..., fmfm
and use them as pipe settings. ii
-th pipe will distribute fifi
units of water per second from junction xixi
to junction yiyi
(if fifi
is negative, then the pipe will distribute |fi||fi|
units of water per second from junction yiyi
to junction xixi
). It is allowed to set fifi
to any integer from −2⋅109−2⋅109
to 2⋅1092⋅109
.
In order for the system to work properly, there are some constraints: for every i∈[1,n]i∈[1,n]
, ii
-th junction has a number sisi
associated with it meaning that the difference between incoming and outcoming flow for ii
-th junction must be exactly sisi
(if sisi
is not negative, then ii
-th junction must receive sisi
units of water per second; if it is negative, then ii
-th junction must transfer |si||si|
units of water per second to other junctions).
Can you choose the integers f1f1
, f2f2
, ..., fmfm
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\n",ans[i]);
74 }
75 }
原文地址：https://www.cnblogs.com/Fang-Hao/p/9255907.html