uva 10746 Crime Wave - The Sequel
2017-02-18 22:58
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原题:
n banks have been robbed this fine day. m (greater than or equal to n) police cruisers are on duty at various locations in the city. n of the cruisers should be dispatched, one to each of the banks, so as to minimize the average time of arrival at the n banks.
Input
The input file contains several sets of inputs. The description of each set is given below:
The first line of input contains 0 < n ≤ m ≤ 20. n lines follow, each containing m positive real numbers: the travel time for cruiser m to reach bank n.
Input is terminated by a case where m = n = 0. This case should not be processed.
Output
For each set of input output a single number: the minimum average travel time, accurate to 2 fractional digits.
Sample Input
3 4
10.0 23.0 30.0 40.0
5.0 20.0 10.0 60.0
18.0 20.0 20.0 30.0
0 0
Sample Output
13.33
中文:
有n个银行被抢劫,有m个警察。给你每个警察到这n银行的时间,现在问你每个银行派一个警察,使得使得总消耗时间的平均和最小是多少?
思路:
裸的最小费用最大流问题,给每个警察到每个银行建立一条边,容量都是1,费用用给定的时间表示。
设置超级源点s’到源点s,容量设置成银行的数量,用来控制出警的数量,然后把s点和所有警察连接上,容量设置为1,费用设置为0,最后把所有银行连接到一个汇点上,每个边的容量都是1,费用都设置为0即可。
此题是小数,所以一定要注意精度转换!用eps最小精度来调节寻找最短增光路的松弛过程!结果也要加上eps
n banks have been robbed this fine day. m (greater than or equal to n) police cruisers are on duty at various locations in the city. n of the cruisers should be dispatched, one to each of the banks, so as to minimize the average time of arrival at the n banks.
Input
The input file contains several sets of inputs. The description of each set is given below:
The first line of input contains 0 < n ≤ m ≤ 20. n lines follow, each containing m positive real numbers: the travel time for cruiser m to reach bank n.
Input is terminated by a case where m = n = 0. This case should not be processed.
Output
For each set of input output a single number: the minimum average travel time, accurate to 2 fractional digits.
Sample Input
3 4
10.0 23.0 30.0 40.0
5.0 20.0 10.0 60.0
18.0 20.0 20.0 30.0
0 0
Sample Output
13.33
中文:
有n个银行被抢劫,有m个警察。给你每个警察到这n银行的时间,现在问你每个银行派一个警察,使得使得总消耗时间的平均和最小是多少?
#include<bits/stdc++.h> using namespace std; const int maxn=101; const double inf=100000000.000; const double eps = 1e-6; struct Edge//边 { int from,to,cap,flow;//出点,入点,容量,当前流量,费用(也就是权值) double cost; Edge(int u,int v,int c,int f,double w):from(u),to(v),cap(c),flow(f),cost(w){} }; struct MCMF { int n,m; vector<Edge> edges;//保存表 vector<int> G[maxn];//保存邻接关系 int inq[maxn];//判断一个点是否在队列当中(SPFA算法当中要用) double d[maxn];//起点到d[i]的最短路径保存值 int p[maxn];//用来记录路径,保存上一条弧 int a[maxn];//找到增广路径后的改进量 void init(int n)//初始化 { this->n=n; for(int i=0;i<=n;i++) G[i].clear(); edges.clear(); } void AddEdge(int from,int to,int cap,double cost)//添加边 { edges.push_back(Edge(from,to,cap,0,cost));//正向 edges.push_back(Edge(to,from,0,0,-cost));//反向 m=edges.size(); G[from].push_back(m-2);//按照边的编号保存邻接关系 G[to].push_back(m-1); } bool BellmanFord(int s,int t,int& flow,double& cost)//最短路径算法 { for(int i=0;i<=n;i++) d[i]=inf*1.0; memset(inq,0,sizeof(inq)); d[s]=0; inq[s]=1; p[s]=0; a[s]=inf; queue<int> Q; Q.push(s); while(!Q.empty()) { int u=Q.front(); Q.pop(); inq[u]=0; for(int i=0;i<G[u].size();i++) { Edge& e=edges[G[u][i]]; if(e.cap>e.flow&&d[e.to]>d[u]+e.cost+eps)//寻找满足容量大于流量的可松弛边 { d[e.to]=d[u]+e.cost; p[e.to]=G[u][i]; a[e.to]=min(a[u],e.cap-e.flow); if(!inq[e.to])//是否在队列当中 { Q.push(e.to); inq[e.to]=1; } } } } if(d[t]>=inf)//如果d[t]没有被更新,相当于没找到增广路径,则没有最大流也没有最小费用 return false; flow+=a[t];//更新最大流 cost+=(double )d[t]*(double)a[t];//单位流量乘以单位路径长度用来计算消耗 for(int u=t;u!=s;u=edges[p[u]].from)//通过使用p[]保存的上一个边的值来对刚刚找到的增广路径上面的流量进行更新 { edges[p[u]].flow+=a[t];//正向变更新 edges[p[u]^1].flow-=a[t];//反向变更新(用位运算实现的) } return true; } int MincostMaxflow(int s,int t,double& cost)//计算从s到t的最小消耗cost,返回最大流 { int flow = 0; cost=0; while(BellmanFord(s,t,flow,cost));//不断寻找最短增广路径,直到找不到为止 return flow; } }; MCMF mcmf; double tmp[21][21]; int main() { int n,m; ios::sync_with_stdio(false); while(cin>>n>>m,n+m) { mcmf.init(100); for(int i=1;i<=n;i++) { for(int j=1;j<=m;j++) { cin>>tmp[i][j]; } } mcmf.AddEdge(0,1,n,0); for(int i=1;i<=m;i++) mcmf.AddEdge(1,i+1,1,0); for(int i=1;i<=m;i++) for(int j=1;j<=n;j++) mcmf.AddEdge(1+i,1+m+j,1,tmp[j][i]); for(int i=1;i<=n;i++) mcmf.AddEdge(m+1+i,m+n+2,1,0); double ans; mcmf.MincostMaxflow(0,m+n+2,ans); cout<<fixed<<setprecision(2)<<(ans/n+eps)<<endl; } return 0; }
思路:
裸的最小费用最大流问题,给每个警察到每个银行建立一条边,容量都是1,费用用给定的时间表示。
设置超级源点s’到源点s,容量设置成银行的数量,用来控制出警的数量,然后把s点和所有警察连接上,容量设置为1,费用设置为0,最后把所有银行连接到一个汇点上,每个边的容量都是1,费用都设置为0即可。
此题是小数,所以一定要注意精度转换!用eps最小精度来调节寻找最短增光路的松弛过程!结果也要加上eps
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