【最小费用最大流模板】【Uva10806+Spring Team PK】Dijkstra, Dijkstra,
题意:从1到n 再从n到1 不经过重复的边 ,(如果是点就是旅行商问题了),问最短路
建立一个超级源S S到1连一条费用为0,容量为2的边,求费用流即可
如果流<2 那么hehe
否则 输出结果
模板来自Kuangbing 如下:
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#include <queue>
#define oo 0x13131313
using namespace std;
const int MAXN=200;
const int MAXM=200000;
const int INF=0x3f3f3f3f;
struct Edge
{
int to,next,cap,flow,cost;
void get(int a,int b,int c,int d)
{
to=a,cap=b,cost=c;next=d;flow=0;
}
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;
void init(int n)
{
N=n;
tol=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
edge[tol].get(v,cap,cost,head[u]);head[u]=tol++;
edge[tol].get(u,0,-cost,head[v]);head[v]=tol++;
}
bool spfa(int s,int t)
{
queue<int>q;
for(int i=0;i<N;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()
{
}
完整代码如下:
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
#include <iostream>
#include <sstream>
#include <string>
#include <queue>
#define oo 0x13131313
using namespace std;
const int MAXN=200;
const int MAXM=200000;
const int INF=0x3f3f3f3f;
struct Edge
{
int to,next,cap,flow,cost;
void get(int a,int b,int c,int d)
{
to=a,cap=b,cost=c;next=d;flow=0;
}
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;
void init(int n)
{
N=n;
tol=0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost)
{
edge[tol].get(v,cap,cost,head[u]);head[u]=tol++;
edge[tol].get(u,0,-cost,head[v]);head[v]=tol++;
}
bool spfa(int s,int t)
{
queue<int>q;
for(int i=0;i<=N;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 NN,MM;
void input()
{
int a,b,c;
for(int i=1;i<=MM;i++)
{
scanf("%d%d%d",&a,&b,&c);
addedge(a,b,1,c);
addedge(b,a,1,c);
}
}
void solve()
{
int ANS=0,t;
addedge(NN+1,1,2,0); //建立源S=NN+1;
t=minCostMaxflow(NN+1,NN,ANS);
if(t==2) printf("%d\n",ANS);
else printf("hehe\n");
}
void File()
{
freopen("a.in","r",stdin);
freopen("a.out","w",stdout);
}
int main()
{
// File();
while(cin>>NN>>MM)
{
init(NN+1);
input();
solve();
}
}
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