Hdu1853/HDU 3488 directed graph, with K circles covering all vertices once // fee stream

This is the Basic Charge flow question type, Split points, create a bipartite graph, and run the maximum flow with the minimum cost. If the maximum stream is N, it indicates that the maximum match is N. All vertices are involved. The inbound and outbound degrees of each vertex are 1, so they are loops.

Hard work is required for weak dishes!

#include<cstdio>#include<iostream>#include<queue>#include<cstring>using namespace std;const int inf=0x3f3f3f3f;int nume=0;int e[50000][4];int head[500];int n,m;void inline  adde(int i,int j,int c,int w){    e[nume][0]=j;e[nume][1]=head[i];head[i]=nume;    e[nume][2]=c;e[nume++][3]=w;    e[nume][0]=i;e[nume][1]=head[j];head[j]=nume;    e[nume][2]=0;e[nume++][3]=-w;}int inq[500];int pre[500];int prv[500];int d[500];bool spfa(int &sum,int &sumflow){    for(int i=0;i<=2*n+2;i++)          {              inq[i]=0;              d[i]=inf;          }    queue<int>q;    q.push(0);    inq[0]=1;    d[0]=0;    while(!q.empty())    {        int cur=q.front();        q.pop();        inq[cur]=0;        for(int i=head[cur];i!=-1;i=e[i][1])        {            int v=e[i][0];            if(e[i][2]>0&&d[cur]+e[i][3]<d[v])            {                d[v]=d[cur]+e[i][3];                pre[v]=i;                prv[v]=cur;                if(!inq[v])                {                    q.push(v);                    inq[v]=1;                }            }        }    }    if(d[2*n+2]==inf)return 0;    int cur=2*n+2;int minf=inf;    while(cur!=0)    {        minf=e[pre[cur]][2]<minf?e[pre[cur]][2]:minf;        cur=prv[cur];    }     cur=2*n+2;    while(cur!=0)    {        e[pre[cur]][2]-=minf;        e[pre[cur]^1][2]+=minf;        cur=prv[cur];    }    sumflow+=minf;    sum+=minf*d[2*n+2];    return 1;}int mincost(int &sumflow){    int sum=0;    while(spfa(sum,sumflow));    return sum;}void init(){    nume=0;    memset(head,-1,sizeof(head));}int main(){    while(~scanf("%d%d",&n,&m))    {        init();        int a,b,c;        for(int i=0;i<m;i++)        {           scanf("%d%d%d",&a,&b,&c);            adde(a,b+n,1,c);        }        for(int i=1;i<=n;i++)        {            adde(0,i,1,0);            adde(i+n,2*n+1,1,0);        }        adde(2*n+1,2*n+2,n,0);        int sumflow=0;        int ans=mincost(sumflow);        if(sumflow!=n)        printf("-1\n");        else        {            printf("%d\n",ans);        }    }    return 0;}