Arrest

Question

问题描述 :

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.
输入:

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
输出:

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
样例输入:

3 4 2
0 1 3
0 2 4
1 3 2
2 3 2
0 0 0
样例输出:

14

Answer 1

 #include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#include<algorithm>
#include<math.h>
#include<queue>
#include<iomanip>
using namespace std;
typedef long long ll;
const   int oo=1e9;
const   int mm=11111111;
const   int mn=888888;
int node,src,dest,edge;
int ver[mm],flow[mm],cost[mm],nex[mm];
int head[mn],dis[mn],p[mn],q[mn],vis[mn];
/**这些变量基本与最大流相同，增加了
 cost 表示边的费用，
 p 记录可行流上节点对应的反向边
 */
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]=0;
    edge=0;
}
void addedge(int u,int v,int f,int c)
{
    ver[edge]=v,flow[edge]=f,cost[edge]=c,nex[edge]=head[u],head[u]=edge++;
    ver[edge]=u,flow[edge]=0,cost[edge]=-c,nex[edge]=head[v],head[v]=edge++;
}
/**以上同最大流*/
/**spfa 求最短路，并用 p 记录最短路上的边*/
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>=mn)?l=0:l)
        for(i=head[u=q[l]],vis[u]=0; i>=0; i=nex[i])
            if(flow[i]&&dis[v=ver[i]]>(tmp=dis[u]+cost[i]))
            {
                dis[v]=tmp;
                p[v]=i^1;
                if(vis[v]) continue;
                vis[q[r++]=v]=1;
                if(r>=mn)r=0;
            }
    return p[dest]>-1;
}
/**源点到汇点的一条最短路即可行流，不断的找这样的可行流*/
int SpfaFlow()
{
    int i,ret=0,delta;
    while(spfa())
    {
        for(i=p[dest],delta=oo; i>=0; i=p[ver[i]])
            if(flow[i^1]<delta)delta=flow[i^1];
        for(i=p[dest]; i>=0; i=p[ver[i]])
            flow[i]+=delta,flow[i^1]-=delta;
        ret+=delta*dis[dest];
    }
    return ret;
}
int tu[110][110];

void floyd(int n)
{
    for(int k=0; k<=n; k++)
        for(int i=0; i<=n; i++)
            for(int j=0; j<=n; j++)
                tu[i][j]=min(tu[i][j],tu[i][k]+tu[k][j]);
}

int main()
{
    int n,m,k;
    while(~scanf("%d%d%d",&n,&m,&k)&&n+m+k)
    {
        int st=2*n+1,ed=2*n+2;
        prepare(2*n+3,st,ed);
        for(int i=0; i<=n; i++)
            for(int j=0; j<=n; j++)
                tu[i][j]=i==j?0:oo;
        while(m--)
        {
            int a,b,c;
            scanf("%d%d%d",&a,&b,&c);
            if(c<tu[a][b])
                tu[a][b]=tu[b][a]=c;
        }
        floyd(n);
        addedge(st,0,k,0);
        addedge(0,ed,k,0);
        for(int i=1; i<=n; i++)
            addedge(0,i,1,tu[0][i]),addedge(i,i+n,1,-100000),addedge(i+n,ed,1,tu[i][0]);
        for(int i=1; i<=n; i++)
            for(int j=i+1; j<=n; j++)
                addedge(i+n,j,1,tu[i][j]);
        printf("%d\n",SpfaFlow()+100000*n);
    }
    return 0;
}