P2153 [SDOI2009]晨跑

思路

典型的最小费用最大流问题，拆点，每个点对应的入点和出点之间连一条cap=1的边表示只能经过一次的限制条件
然后其他边从u的出点连向v的入点即可

代码

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
using namespace std;
const int MAXN = 1000;
const int INF = 0x3f3f3f3f;
struct Edge{
    int u,v,cap,cost,flow;
};
int n,m;
vector<Edge> edges;
vector<int> G[MAXN];
void addedge(int u,int v,int cap,int cost){
    edges.push_back((Edge){u,v,cap,cost,0});
    edges.push_back((Edge){v,u,0,-cost,0});
    int cnt=edges.size();
    G[u].push_back(cnt-2);
    G[v].push_back(cnt-1);
}
int a[MAXN],d[MAXN],p[MAXN],vis[MAXN],s,t;
queue<int> q;
bool spfa(int &flow,int &cost){
    memset(d,0x3f,sizeof(d));
    memset(p,0,sizeof(p));
    memset(vis,0,sizeof(vis));
    q.push(s);
    d[s]=0;
    p[s]=0;
    a[s]=INF;
    vis[s]=true;
    while(!q.empty()){
        int x=q.front();
        q.pop();
        vis[x]=false;
        for(int i=0;i<G[x].size();i++){
            Edge &e = edges[G[x][i]];
            if(e.cap>e.flow&&d[x]+e.cost<d[e.v]){
                d[e.v]=d[x]+e.cost;
                a[e.v]=min(a[x],e.cap-e.flow);
                p[e.v]=G[x][i];
                if(!vis[e.v]){
                    vis[e.v]=true;
                    q.push(e.v);
                }
            }
        }
    }
    if(d[t]==INF)
        return false;
    flow+=a[t];
    cost+=a[t]*d[t];
    for(int i=t;i!=s;i=edges[p[i]].u){
        edges[p[i]].flow+=a[t];
        edges[p[i]^1].flow-=a[t];
    }
    return true;
}
void MCMF(int &flow,int &cost){
    flow=cost=0;
    while(spfa(flow,cost));
}
int main(){
    scanf("%d %d",&n,&m);
    s=1,t=n;
    addedge(1,n+1,INF,0);
    for(int i=2;i<n;i++)
        addedge(i,n+i,1,0);
    addedge(n,2*n,INF,0);
    for(int i=1;i<=m;i++){
        int a,b,c;
        scanf("%d %d %d",&a,&b,&c);
        addedge(n+a,b,1,c);
    }
    int flow,cost;
    MCMF(flow,cost);
    printf("%d %d
",flow,cost);
    return 0;
}