HDU 4411 Arrest（Floyd+最小费用最大流）

Arrest

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1932    Accepted Submission(s): 765

Problem Description

There are (N+1) cities on TAT island. City 0 is where police headquarter located. The economy of other cities numbered from 1 to N ruined these years because they are all controlled by mafia. The police plan to catch all the mafia gangs in these N cities all over the year, and they want to succeed in a single mission. They figure out that every city except city 0 lives a mafia gang, and these gangs have a simple urgent message network: if the gang in city i (i>1) is captured, it will send an urgent message to the gang in city i-1 and the gang in city i -1 will get the message immediately.
The mission must be carried out very carefully. Once a gang received an urgent message, the mission will be claimed failed.
You are given the map of TAT island which is an undirected graph. The node on the graph represents a city, and the weighted edge represents a road between two cities(the weight means the length). Police headquarter has sent k squads to arrest all the mafia gangs in the rest N cities. When a squad passes a city, it can choose to arrest the gang in the city or to do nothing. These squads should return to city 0 after the arrest mission.
You should ensure the mission to be successful, and then minimize the total length of these squads traveled.

 

Input

There are multiple test cases.
Each test case begins with a line with three integers N, M and k, here M is the number of roads among N+1 cities. Then, there are M lines. Each of these lines contains three integers X, Y, Len, which represents a Len kilometer road between city X and city Y. Those cities including city 0 are connected.
The input is ended by “0 0 0”.
Restrictions: 1 ≤ N ≤ 100, 1 ≤ M ≤ 4000, 1 ≤ k ≤ 25, 0 ≤ Len ≤ 1000

 

Output

For each test case,output a single line with a single integer that represents the minimum total length of these squads traveled.

 

Sample Input

3 4 20 1 30 2 41 3 22 3 20 0 0

 

Sample Output

14

 

Source

2012 ACM/ICPC Asia Regional Hangzhou Online

 

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题目大意：

    有一个城市0，和N个城市1～N，给出一些城市之间的距离。城市0最多有K个警察，要按照城市从1到N的顺序抓住N个城市中每个城市的罪犯再返回城市0，求最小路程。

解题思路：

    首先，最小费用最大流求的是在一个有流量限制的网络中，保证流量最大的前提下使得花费最小的流量与花费。

    那么我们很容易就可以想到把警察个数看作流量，距离看作花费。由于不一定要使用全部警察。我们只需要在在城市0和终点之间连一个容量无限，花费为0的边即可。其它的就是很普通的拆点，流量限制，具体看代码。

AC代码：

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <vector>#include <queue>using namespace std;#define mem(a,b) memset((a),(b),sizeof(a))#define INF 0x01010101#define fi first#define se secondstruct Edge{    int to,next,cap,flow,cost;};const int maxn=100+3;const int maxv=maxn*2;int dist[maxn][maxn];int N,M,K,V;int head[maxv],tol;int pre[maxv],dis[maxv];bool vis[maxv];Edge edge[maxv*maxv];void init(){    tol=0;    mem(head,-1);}void add_edge(int u,int v,int cap,int cost){    edge[tol].to=v;    edge[tol].cap=cap;    edge[tol].cost=cost;    edge[tol].flow=0;    edge[tol].next=head[u];    head[u]=tol++;    edge[tol].to=u;    edge[tol].cost=-cost;    edge[tol].flow=0;    edge[tol].next=head[v];    head[v]=tol++;}bool spfa(int s,int t){    queue<int> q;    for(int i=0;i<V;++i)    {        dis[i]=INF;        vis[i]=false;        pre[i]=-1;    }    dis[s]=0;    vis[s]=true;    q.push(s);    while(!q.empty())    {        int u=q.front(); q.pop();        vis[u]=false;        for(int i=head[u];i!=-1;i=edge[i].next)        {            int v=edge[i].to;            if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost)            {                dis[v]=dis[u]+edge[i].cost;                pre[v]=i;                if(!vis[v])                {                    vis[v]=true;                    q.push(v);                }            }        }    }    if(pre[t]==-1)        return false;    else return true;}int minCostMaxFlow(int s,int t,int &cost){    int flow=0;    cost=0;    while(spfa(s,t))    {        int Min=INF;        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])            if(Min>edge[i].cap-edge[i].flow)                Min=edge[i].cap-edge[i].flow;        for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])        {            edge[i].flow+=Min;            edge[i^1].flow-=Min;            cost+=edge[i].cost*Min;        }        flow+=Min;    }    return flow;}int main(){    while(~scanf("%d%d%d",&N,&M,&K)&&(N||M||K))    {        //0~N城市        //N+1~N*2抓强盗后的城市        //N*2+1源点        //N*2+2汇点        V=N*2+3;        init();        int s=N*2+1,t=N*2+2;        mem(dist,0x01);        for(int i=0;i<M;++i)        {            int a,b,c;            scanf("%d%d%d",&a,&b,&c);            dist[a][b]=dist[b][a]=min(dist[a][b],c);//有重边        }        for(int i=0;i<=N;++i)            dist[i][i]=0;        for(int k=0;k<=N;++k)//Floyd-Warshall算法，求任何两点之间的最短距离            for(int i=0;i<=N;++i)                for(int j=0;j<=N;++j)                    dist[i][j]=min(dist[i][j],dist[i][k]+dist[k][j]);        add_edge(s,0,K,0);//连接超级源点和城市0        add_edge(0,t,INF,0);//不一定使用全部警察        for(int i=1;i<=N;++i)        {            add_edge(i,i+N,INF-1,0);//最小流量限制为1            add_edge(i,i+N,1,-INF);            add_edge(0,i,INF,dist[0][i]);//警察可以直接从城市0过来            add_edge(i+N,t,INF,dist[i][0]);//警察回到城市0            for(int j=1;j<i;++j)//和前面的每个城市相连                add_edge(j+N,i,INF,dist[j][i]);        }        int ans;        minCostMaxFlow(s,t,ans);        printf("%d\n",ans+N*INF);    }        return 0;}