[BZOJ1927][SDOI2010]星际竞速 最小费用最大流

用流量限制次数，用费用限制距离。
这一题和最小路径覆盖很像，所有边容量为1，把每个点拆成u，u′，原图上u，v连成u，v′，费用为距离，每个u′连向汇点，费用0，源点连向每个u，费用0，源点连向每个u′，费用为瞬移代价，直接跑最小费用最大流就可以了。
注意最小费用最大流每个点记的是来边而不是来点。

#include<ctime>#include<cstdio>#include<cstdlib>#include<cstring>#include<algorithm>#include<iostream>#include<string>#include<cassert>#include<cmath>#include<vector>#include<queue>#include<stack>#include<map>#include<sstream>#include<climits>#define X first#define Y second#define DB double#define lc now<<1#define rc now<<1|1#define MP make_pair#define LL long long#define pb push_back#define sqr(_) ((_)*(_))#define INF 0x3f3f3f3f#define pii pair<int,int>#define pdd pair<DB,DB>#define ull unsigned LL#define DEBUG(...) fprintf(stderr,__VA_ARGS__)using namespace std;const int MAXN=2000;template<typename T>void Read(T& x){    x=0;int flag=0,sgn=1;char c;    while(c=getchar())    {        if(c=='-')sgn=-1;        else if(c>='0'&&c<='9')x*=10,x+=c-'0',flag=1;        else if(flag)break;    }    x*=sgn;}int w[MAXN],n,m;struct Edge{    int u,v,cap,flow,cost;    Edge(int u=0,int v=0,int cap=0,int flow=0,int cost=0):        u(u),v(v),cap(cap),flow(flow),cost(cost){}};struct Mincost{    vector<int> G[MAXN];    vector<Edge> edges;    int p[MAXN],inq[MAXN],d[MAXN],a[MAXN],S,T;    void add(int u,int v,int cap,int cost)    {        edges.pb(Edge(u,v,cap,0,cost));        edges.pb(Edge(v,u,0,0,-cost));        int k=edges.size();        G[u].pb(k-2);        G[v].pb(k-1);    }    void init()    {        memset(p,0,sizeof(p));        memset(inq,0,sizeof(inq));        memset(a,0,sizeof(a));        memset(d,INF,sizeof(d));    }    bool SPFA(int& flow,int& cost)    {        init();        queue<int> q;        q.push(S);        inq[S]=1;        d[S]=0;        a[S]=INF;        while(!q.empty())        {            int u=q.front();q.pop();            inq[u]=0;            for(int i=0;i<G[u].size();i++)            {                Edge& e=edges[G[u][i]];                if(d[e.v]>d[u]+e.cost&&e.cap>e.flow)                {                    d[e.v]=d[u]+e.cost;                    p[e.v]=G[u][i];                    a[e.v]=min(a[u],e.cap-e.flow);                    if(!inq[e.v])                    {                        q.push(e.v);                        inq[e.v]=1;                    }                }            }        }        if(d[T]==INF)return 0;        flow+=a[T];        cost+=d[T]*a[T];        int u=T;        while(u!=S)        {            edges[p[u]].flow+=a[T];            edges[p[u]^1].flow-=a[T];            u=edges[p[u]].u;        }        return 1;    }    int mincost()    {        int flow=0,cost=0;        while(SPFA(flow,cost));        return cost;    }}Graph;int main(){#ifndef ONLINE_JUDGE    freopen("speed.in","r",stdin);    freopen("speed.out","w",stdout);#endif    Read(n);Read(m);    for(int i=1;i<=n;i++)        Read(w[i]);    for(int i=1;i<=m;i++)    {        int a,b,c;        Read(a),Read(b),Read(c);        if(a>b)swap(a,b);        Graph.add(a,b+n,1,c);    }    Graph.S=2*n+1,Graph.T=2*n+2;    for(int i=1;i<=n;i++)    {        Graph.add(Graph.S,i,1,0);        Graph.add(i+n,Graph.T,1,0);        Graph.add(Graph.S,i+n,1,w[i]);    }    cout<<Graph.mincost()<<endl;}