poj1459 Power Network
2015-12-08 00:18
260 查看
Power Network
Description
A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= pmax(u) of power, may consume an amount
0 <= c(u) <= min(s(u),cmax(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power
transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= lmax(u,v) of power delivered by u to v. Let Con=Σuc(u) be the power consumed in the net. The problem is to compute the maximum value of
Con.
![](http://poj.org/images/1459_1.jpg)
An example is in figure 1. The label x/y of power station u shows that p(u)=x and pmax(u)=y. The label x/y of consumer u shows that c(u)=x and cmax(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and lmax(u,v)=y.
The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.
Input
There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets
(u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of lmax(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of pmax(u). The data set
ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of cmax(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can
occur freely in input. Input data terminate with an end of file and are correct.
Output
For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.
Sample Input
Sample Output
Hint
The sample input contains two data sets. The first data set encodes a network with 2 nodes, power station 0 with pmax(0)=15 and consumer 1 with cmax(1)=20, and 2 power transport lines with lmax(0,1)=20 and lmax(1,0)=10. The maximum value of Con is 15. The second
data set encodes the network from figure 1.
Source
Southeastern Europe 2003
最大流模板。
题目要求多源多汇最大流。方法为添加两个点,分别为超级源点S和超级汇点T,再将所有源点与S连边,所有汇点与T连边,求S到T的最大流。
注意:BFS用到的STL队列模板要写到子函数里,具体为什么我也不太清楚,写在外面会出错。
Time Limit: 2000MS | Memory Limit: 32768K | |
Total Submissions: 25843 | Accepted: 13488 |
A power network consists of nodes (power stations, consumers and dispatchers) connected by power transport lines. A node u may be supplied with an amount s(u) >= 0 of power, may produce an amount 0 <= p(u) <= pmax(u) of power, may consume an amount
0 <= c(u) <= min(s(u),cmax(u)) of power, and may deliver an amount d(u)=s(u)+p(u)-c(u) of power. The following restrictions apply: c(u)=0 for any power station, p(u)=0 for any consumer, and p(u)=c(u)=0 for any dispatcher. There is at most one power
transport line (u,v) from a node u to a node v in the net; it transports an amount 0 <= l(u,v) <= lmax(u,v) of power delivered by u to v. Let Con=Σuc(u) be the power consumed in the net. The problem is to compute the maximum value of
Con.
![](http://poj.org/images/1459_1.jpg)
An example is in figure 1. The label x/y of power station u shows that p(u)=x and pmax(u)=y. The label x/y of consumer u shows that c(u)=x and cmax(u)=y. The label x/y of power transport line (u,v) shows that l(u,v)=x and lmax(u,v)=y.
The power consumed is Con=6. Notice that there are other possible states of the network but the value of Con cannot exceed 6.
Input
There are several data sets in the input. Each data set encodes a power network. It starts with four integers: 0 <= n <= 100 (nodes), 0 <= np <= n (power stations), 0 <= nc <= n (consumers), and 0 <= m <= n^2 (power transport lines). Follow m data triplets
(u,v)z, where u and v are node identifiers (starting from 0) and 0 <= z <= 1000 is the value of lmax(u,v). Follow np doublets (u)z, where u is the identifier of a power station and 0 <= z <= 10000 is the value of pmax(u). The data set
ends with nc doublets (u)z, where u is the identifier of a consumer and 0 <= z <= 10000 is the value of cmax(u). All input numbers are integers. Except the (u,v)z triplets and the (u)z doublets, which do not contain white spaces, white spaces can
occur freely in input. Input data terminate with an end of file and are correct.
Output
For each data set from the input, the program prints on the standard output the maximum amount of power that can be consumed in the corresponding network. Each result has an integral value and is printed from the beginning of a separate line.
Sample Input
2 1 1 2 (0,1)20 (1,0)10 (0)15 (1)20 7 2 3 13 (0,0)1 (0,1)2 (0,2)5 (1,0)1 (1,2)8 (2,3)1 (2,4)7 (3,5)2 (3,6)5 (4,2)7 (4,3)5 (4,5)1 (6,0)5 (0)5 (1)2 (3)2 (4)1 (5)4
Sample Output
15 6
Hint
The sample input contains two data sets. The first data set encodes a network with 2 nodes, power station 0 with pmax(0)=15 and consumer 1 with cmax(1)=20, and 2 power transport lines with lmax(0,1)=20 and lmax(1,0)=10. The maximum value of Con is 15. The second
data set encodes the network from figure 1.
Source
Southeastern Europe 2003
最大流模板。
题目要求多源多汇最大流。方法为添加两个点,分别为超级源点S和超级汇点T,再将所有源点与S连边,所有汇点与T连边,求S到T的最大流。
注意:BFS用到的STL队列模板要写到子函数里,具体为什么我也不太清楚,写在外面会出错。
#include<iostream> #include<cstdio> #include<algorithm> #include<cstdlib> #include<cmath> #include<cstring> #include<queue> #define F(i,j,n) for(int i=j;i<=n;i++) #define D(i,j,n) for(int i=j;i>=n;i--) #define LL long long #define pa pair<int,int> #define MAXN 105 #define MAXM 30000 #define INF 1000000000 using namespace std; struct edge_type { int next,to,v; }e[MAXM]; int n,m,np,nc,x,y,z,s,t,ans,cnt,dis[MAXN],head[MAXN],cur[MAXN]; inline void add_edge(int x,int y,int v) { e[++cnt]=(edge_type){head[x],y,v};head[x]=cnt; e[++cnt]=(edge_type){head[y],x,0};head[y]=cnt; } inline bool bfs() { queue<int>q; memset(dis,-1,sizeof(dis)); dis[s]=0;q.push(s); while (!q.empty()) { int tmp=q.front();q.pop(); if (tmp==t) return true; for(int i=head[tmp];i;i=e[i].next) if (e[i].v&&dis[e[i].to]==-1) { dis[e[i].to]=dis[tmp]+1; q.push(e[i].to); } } return false; } inline int dfs(int x,int f) { int tmp,sum=0; if (x==t) return f; for(int &i=cur[x];i;i=e[i].next) { int y=e[i].to; if (e[i].v&&dis[y]==dis[x]+1) { tmp=dfs(y,min(f-sum,e[i].v)); e[i].v-=tmp;e[i^1].v+=tmp;sum+=tmp; if (sum==f) return sum; } } if (!sum) dis[x]=-1; return sum; } inline void dinic() { ans=0; while (bfs()) { if (!dis[t]) return; F(i,1,n+2) cur[i]=head[i]; ans+=dfs(s,INF); } } int main() { while (scanf("%d%d%d%d",&n,&np,&nc,&m)!=EOF) { char ch; s=n+1;t=n+2;cnt=1; memset(head,0,sizeof(head)); F(i,1,m) { ch=getchar(); while (ch!='(') ch=getchar(); scanf("%d,%d)%d",&x,&y,&z);x++;y++; if (x!=y) add_edge(x,y,z); } F(i,1,np) { ch=getchar(); while (ch!='(') ch=getchar(); scanf("%d)%d",&x,&z);x++; add_edge(s,x,z); } F(i,1,nc) { ch=getchar(); while (ch!='(') ch=getchar(); scanf("%d)%d",&x,&z);x++; add_edge(x,t,z); } dinic(); printf("%d\n",ans); } }
相关文章推荐
- apk更新、下载、安装(三)---DownloadManager ui方式【有卡顿bug】
- C#单例模式的泛型定义
- android真机调试看不到logcat信息
- Swift中的元组(turple)
- 方法的参数传递机制
- 基于HTML5的WebGL结合Box2DJS物理引擎应用
- python 批量下载图片
- 实验报告(实验四)
- 深夜喂一口鸡汤
- poj-2342Anniversary party
- JS字符串false转boolean
- 单词识别 c++ primer plus 第六章变成练习 第7题
- linux进程间的网络通信
- 学习数据结构好的博客
- .NET AutoCAD二次开发之路(二、直线篇)
- 红黑树
- 10007---AngularJS 控制器
- HP ProLiant BL460c Gen8 服务器部署手记
- linux SVN服务器创建版本库
- c++ primer plus 第六章变成练习 第7题