hdu 4411 最小费用最大流
http://acm.hdu.edu.cn/showproblem.php?pid=4411
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.
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
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 2 0 1 3 0 2 4 1 3 2 2 3 2 0 0 0
Sample Output
14
/**
hdu 4411 最小费用最大流
题目大意：给定一个图，有0~n个共n+1个顶点，从0点出发的k个警察去其余n个点抓小偷，最后还要回到0点（可以有的警察不出动），必须保证的是在抓第i个顶点
的小偷是第i-1顶点的小偷已经被抓住了，否则抓捕计划失败。问怎么样做能使所有警察走的路程最短
解题思路：很考思维的一道网络流的题，建图的难点在于抓第i个顶点的小偷的时候必须保证第i-1点的小偷已经被抓走了和警察不一定全用。对于点i分成两个点i1，i2中间用流量为1权值为-100000的边
连接，然后对i2点和j1(j>i)的点连一条流量为1，权值为ij的最短距离的边就可以了。源点到0连流量为k费用为0的点每个i1点和0之间连权值为1，费用为0i的
最短距离，每个i2点和汇点连流量为1，费用为0i最小距离的点。由于k个警察不一定全用到，那么在0和汇点之间连一条流量为k，费用为0的点。
最后最消费用加上k*10000就是答案了。
注：为什么连汇点和0之间的边，我这样理解我们每个i1和i2之间都有费用为-100000的点也就是i1和i2之间还能连的话，一定不用走0~汇点的边，这样就不会改变最小费用了
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <string.h>
using namespace std;
const int oo=1e9;
const int maxm=1111111;
const int maxn=2222;
int node,src,dest,edge;
int head[maxn],p[maxn],dis[maxn],q[maxn],vis[maxn];
struct edgenode
{
int to;
int flow;
int cost;
int next;
} edges[maxm];
void prepare(int _node,int _src,int _dest);
void addedge(int u,int v,int f,int c);
bool spfa();
inline int min(int a,int b)
{
return a<b?a:b;
}
inline void prepare(int _node,int _src,int _dest)
{
node=_node;
src=_src;
dest=_dest;
for (int i=0; i<node; i++)
{
head[i]=-1;
vis[i]=false;
}
edge=0;
}
void addedge(int u,int v,int f,int c)
{
edges[edge].flow=f;
edges[edge].cost=c;
edges[edge].to=v;
edges[edge].next=head[u];
head[u]=edge++;
edges[edge].flow=0;
edges[edge].cost=-c;
edges[edge].to=u;
edges[edge].next=head[v];
head[v]=edge++;
}
bool spfa()
{
int i,u,v,l,r=0,tmp;
for (i=0; i<node; i++) dis[i]=oo;
dis[q[r++]=src]=0;
p[src]=p[dest]=-1;
for (l=0; l!=r; ((++l>=maxn)?l=0:1))
{
for (i=head[u=q[l]],vis[u]=false; i!=-1; i=edges[i].next)
{
if (edges[i].flow&&dis[v=edges[i].to]>(tmp=dis[u]+edges[i].cost))
{
dis[v]=tmp;
p[v]=i^1;
if (vis[v]) continue;
vis[q[r++]=v]=true;
if (r>=maxn) r=0;
}
}
}
return p[dest]>=0;
}
int spfaflow()
{
int i,ret=0,delta;
while (spfa())
{
for (i=p[dest],delta=oo; i>=0; i=p[edges[i].to])
{
delta=min(delta,edges[i^1].flow);
}
for (int i=p[dest]; i>=0; i=p[edges[i].to])
{
edges[i].flow+=delta;
edges[i^1].flow-=delta;
}
ret+=delta*dis[dest];
}
return ret;
}
int n,m,k;
int a[105][105];
int main()
{
while(~scanf("%d%d%d",&n,&m,&k))
{
if(n==0&&m==0&&k==0)break;
memset(a,0x3f3f3f3f,sizeof(a));
for(int i=0; i<m; i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
if(a[u][v]>w)
a[u][v]=a[v][u]=w;
}
for(int k=0; k<=n; k++)
{
for(int i=0; i<=n; i++)
{
for(int j=0; j<=n; j++)
{
a[i][j]=min(a[i][j],a[i][k]+a[k][j]);
}
}
}
prepare(2*n+3,2*n+1,2*n+2);
addedge(src,0,k,0);
addedge(0,dest,k,0);
for(int i=1;i<=n;i++)
{
addedge(0,i,1,a[0][i]);
addedge(i,i+n,1,-100000);
addedge(i+n,dest,1,a[0][i]);
}
for(int i=1;i<=n;i++)
{
for(int j=i+1;j<=n;j++)
{
addedge(i+n,j,1,a[i][j]);
}
}
printf("%d\n",spfaflow()+n*100000);
}
return 0;
}