uva 1349 Optimal Bus Route Design(拆点,费用流)
题意:给出\(n(1 \leq n \leq 100)\)个点,每个点与其他一些点相连,边上有权值,现在需要选一些边将所有点包括在一个环内(有且仅有一个环包含每个点,可以有多个环),使得所选边权值之和最小。
题解:对于每个点存在唯一环包括他,等价于每个点存在唯一后继,对于唯一性,可以想到二分图,将原来每个点拆为两个点,然后加个源和汇s,t,在将边权转化为费用跑费用流就行了。
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<vector>
#include<queue>
#include<stack>
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define pb push_back
#define fi first
#define se second
typedef vector<int> VI;
typedef long long ll;
typedef pair<int,int> PII;
const int inf=0x3fffffff;
const ll mod=1000000007;
const int MAXN =200+10;
const int MAXM = 21000;
const int INF = 1e9;
struct Edge
{
int to,next,cap,flow,cost;
}edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数,节点编号从0~N-1
void addedge(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].cap = 0;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 < 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;
}
//返回的是最大流,cost存的是最小费用
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()
{
int n;
while(~scanf("%d",&n)&&n)
{
int st=0,ed=2*n+1;
N=ed+1;
tol=0;
rep(i,0,ed+1) head[i]=-1;
rep(i,1,n+1)
{
int p;
while(~scanf("%d",&p))
{
if(!p) break;
int w;
scanf("%d",&w);
addedge(i,n+p,1,w);
}
}
rep(i,1,n+1) addedge(st,i,1,0),addedge(n+i,ed,1,0);
int cost=0;
int t=minCostMaxflow(st,ed,cost);
if(t<n) puts("N");
else printf("%d\n",cost);
}
return 0;
}