lightoj--1155-- Power Transmission (最大流拆点)
2015-12-13 19:44
369 查看
Power Transmission
Submit
Status
Description
DESA is taking a new project to transfer power. Power is generated by the newly established plant in Barisal. The main aim of this project is to transfer Power in Dhaka. As Dhaka is a megacity with almost 10 million people DESA wants to transfer maximum
amount of power through the network. But as always occurs in case of power transmission it is tough to resist loss. So they want to use some regulators whose main aims are to divert power through several outlets without any loss.
Each such regulator has different capacity. It means if a regulator gets 100 units of power and its capacity is 80 units then remaining 20 units of power will be lost. Moreover each unidirectional link (connectors among regulators) has a certain capacity.
A link with capacity 20 units cannot transfer power more than 20 units. Each regulator can distribute the input power among the outgoing links so that no link capacity is over flown. DESA wants to know the maximum amount of power which can be transmitted throughout
the network so that no power loss occurs. That is the job you have to do.
(Do not try to mix the above description with the real power transmission.)
Input
Input starts with an integer T (≤ 50), denoting the number of test cases.
The input will start with a positive integer N (1 ≤ N ≤ 100) indicates the number of regulators. The next line contains
N positive integers indicating the capacity of each regulator from
1 to N. All the given capacities will be positive and not greater than
1000. The next line contains another positive integer M which is the number of links available among the regulators. Each of the following
M lines contains three positive integers i j C.
'i' and 'j' are the regulator index (1 ≤ i, j ≤ N, i ≠ j, 1 ≤ C ≤ 1000) and
C is the capacity of the link. Power can be transferred from
ith regulator to jth regulator. From a regulator
i to another regulator j, there can be at most one link.
The next line contains two positive integers B and D (1 ≤ B, D and B + D ≤ N).
B is the number of regulators which are the entry point of the network. Power generated in Barisal must enter in the network through these entry points. Similarly
D is the number of regulators connected to Dhaka. These links are special and have infinite capacity. Next line will contain
B+D integers each of which is an index of regulator. The first
B integers are the index of regulators connected with Barisal. Regulators connected with Barisal are not connected with Dhaka.
Output
For each case of input, print the case number and the maximum amount of power which can be transferred from Barisal to Dhaka.
Sample Input
2
4
10 20 30 40
6
1 2 5
1 3 10
1 4 13
2 3 5
2 4 7
3 4 20
3 1
1 2 3 4
2
50 100
1
1 2 100
1 1
1 2
Sample Output
Case 1: 37
Case 2: 50
Source
Problem Setter: Md. Kamruzzaman
Special Thanks: Jane Alam Jan (Solution, Dataset)
Time Limit: 2000MS | Memory Limit: 32768KB | 64bit IO Format: %lld & %llu |
Status
Description
DESA is taking a new project to transfer power. Power is generated by the newly established plant in Barisal. The main aim of this project is to transfer Power in Dhaka. As Dhaka is a megacity with almost 10 million people DESA wants to transfer maximum
amount of power through the network. But as always occurs in case of power transmission it is tough to resist loss. So they want to use some regulators whose main aims are to divert power through several outlets without any loss.
Each such regulator has different capacity. It means if a regulator gets 100 units of power and its capacity is 80 units then remaining 20 units of power will be lost. Moreover each unidirectional link (connectors among regulators) has a certain capacity.
A link with capacity 20 units cannot transfer power more than 20 units. Each regulator can distribute the input power among the outgoing links so that no link capacity is over flown. DESA wants to know the maximum amount of power which can be transmitted throughout
the network so that no power loss occurs. That is the job you have to do.
(Do not try to mix the above description with the real power transmission.)
Input
Input starts with an integer T (≤ 50), denoting the number of test cases.
The input will start with a positive integer N (1 ≤ N ≤ 100) indicates the number of regulators. The next line contains
N positive integers indicating the capacity of each regulator from
1 to N. All the given capacities will be positive and not greater than
1000. The next line contains another positive integer M which is the number of links available among the regulators. Each of the following
M lines contains three positive integers i j C.
'i' and 'j' are the regulator index (1 ≤ i, j ≤ N, i ≠ j, 1 ≤ C ≤ 1000) and
C is the capacity of the link. Power can be transferred from
ith regulator to jth regulator. From a regulator
i to another regulator j, there can be at most one link.
The next line contains two positive integers B and D (1 ≤ B, D and B + D ≤ N).
B is the number of regulators which are the entry point of the network. Power generated in Barisal must enter in the network through these entry points. Similarly
D is the number of regulators connected to Dhaka. These links are special and have infinite capacity. Next line will contain
B+D integers each of which is an index of regulator. The first
B integers are the index of regulators connected with Barisal. Regulators connected with Barisal are not connected with Dhaka.
Output
For each case of input, print the case number and the maximum amount of power which can be transferred from Barisal to Dhaka.
Sample Input
2
4
10 20 30 40
6
1 2 5
1 3 10
1 4 13
2 3 5
2 4 7
3 4 20
3 1
1 2 3 4
2
50 100
1
1 2 100
1 1
1 2
Sample Output
Case 1: 37
Case 2: 50
Source
Problem Setter: Md. Kamruzzaman
Special Thanks: Jane Alam Jan (Solution, Dataset)
#include<stdio.h> #include<string.h> #include<queue> #include<stack> #include<algorithm> using namespace std; #define MAXM 300 #define MAXN 50000 #define INF 0x3f3f3f int dis[MAXM],vis[MAXM],cur[MAXM],head[MAXM]; int n,top,need[MAXM]; struct node { int u,v,cap,flow,next; }edge[MAXN]; void init() { top=0; memset(head,-1,sizeof(head)); } void add(int a,int b,int c) { node E1={a,b,c,0,head[a]}; edge[top]=E1; head[a]=top++; node E2={b,a,0,0,head[b]}; edge[top]=E2; head[b]=top++; } void getmap() { scanf("%d",&n); memset(need,0,sizeof(need)); for(int i=1;i<=n;i++) { scanf("%d",&need[i]); add(i,i+n,need[i]); } int m; scanf("%d",&m); while(m--) { int a,b,c; scanf("%d%d%d",&a,&b,&c); add(a+n,b,c); } int b,d; scanf("%d%d",&b,&d); while(b--) { int c; scanf("%d",&c); add(0,c,need[c]); } while(d--) { int c; scanf("%d",&c); add(c+n,2*n+1,need[c]); } } bool bfs(int s,int t) { queue<int>q; memset(vis,0,sizeof(vis)); memset(dis,-1,sizeof(dis)); q.push(s); vis[s]=1; dis[s]=0; while(!q.empty()) { int u=q.front(); q.pop(); for(int i=head[u];i!=-1;i=edge[i].next) { node E=edge[i]; if(E.cap>E.flow&&!vis[E.v]) { vis[E.v]=1; dis[E.v]=dis[E.u]+1; if(E.v==t) return true; q.push(E.v); } } } return false; } int dfs(int x,int a,int e) { if(x==e||a==0) return a; int flow=0,f; for(int i=cur[x];i!=-1;i=edge[i].next) { node &E=edge[i]; if(dis[E.v]==dis[E.u]+1&&(f=dfs(E.v,min(a,E.cap-E.flow),e))>0) { E.flow+=f; edge[i^1].flow-=f; a-=f; flow+=f; if(a==0) break; } } return flow; } int MAXflow(int s,int t) { int flow=0; while(bfs(s,t)) { memcpy(cur,head,sizeof(head)); flow+=dfs(s,INF,t); } return flow; } int main() { int t; int k=1; scanf("%d",&t); while(t--) { init(); getmap(); printf("Case %d: %d\n",k++,MAXflow(0,2*n+1)); } return 0; }
相关文章推荐
- 20135218——信息安全系统设计第十四周学习总结
- Android基础(一)
- mac上一键配置和安装adb驱动或者环境
- Oracle计算两个整数的和与这两个整数的差与商
- java web mysql数据库插入数据乱码问题解决方法
- Js中Prototype、__proto__、Constructor、Object、Function关系介绍
- lua语言基础
- java 线程技术详解
- c++_6 : 构造函数和析构函数
- LeetCode Longest Substring Without Repeating Characters
- IOS视频开发之MPMoviePlayerController
- gdb用法用例与问题解答
- gdb用法用例与问题解答
- next数组的理解
- ARM_s5pv210_porting_1
- Android 上拉加载 PullToRefresh
- %lf\n和%f\n的区别
- 简单逆向分析使用案例(4)--CrackMe_03.exe 修改bug
- AutoCAD二次开发三种添加插件按钮的方法之一
- gem5: the trace of cache set and cache line write hit counters