HDU 1853 Cyclic Tour

每个点在且仅在一个有向圈内，所以出读入度卡死为一，裂点为出度控制点和入度控制点……

#include <algorithm>
#include <iostream>
#include <fstream>
#include <sstream>
#include <iomanip>

#include <map>
#include <set>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <vector>
#include <string>
#include <bitset>
#include <memory>
#include <complex>
#include <numeric>

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <math.h>
#include <time.h>
#include <ctype.h>
#include <locale.h>

using namespace std;

#pragma pack(4)

const double  eps = 1e-8;
const double   pi = acos(-1.0);
const int     inf = 0x7f7f7f7f;
const __int64 INF = 0x7f7f7f7f7f7f7f7fll;

#define   at(a,i) ((a)&(1<<(i)))
#define   nt(a,i) ((a)^(1<<(i)))
#define set1(a,i) ((a)|(1<<(i)))
#define set0(a,i) ((a)&(~(1<<(i))))

#define gret(a,b) (((a)-(b))>eps)
#define less(a,b) (((b)-(a))>eps)
#define greq(a,b) (((a)-(b))>-eps)
#define leeq(a,b) (((b)-(a))>-eps)
#define equl(a,b) (fabs((a)-(b))<eps)

#define lmax(a,b) ((a)>(b)?(a):(b))
#define lmin(a,b) ((a)<(b)?(a):(b))
#define fmax(a,b) (gret(a,b)?(a):(b))
#define fmin(a,b) (less(a,b)?(a):(b))

const int MAXV = 240;
const int MAXE = 24000;

struct node
{
    int v,w,c;
    node(int _v=0,int _w=0,int _c=0):
        v(_v),w(_w),c(_c){}
    bool operator < (const node argu) const
    {
        return w<argu.w;
    }
}G[MAXE];
int _index,pre[MAXV],next[MAXE];

void clear(void)
{
    _index=0;
    memset(pre,-1,sizeof(pre));
}
void add(int u,int v,int w=0,int c=0)
{
    G[_index].v=v;
    G[_index].w=w;
    G[_index].c=c;
    next[_index]=pre[u];
    pre[u]=_index++;

    G[_index].v=u;
    G[_index].w=0;
    G[_index].c=-c;
    next[_index]=pre[v];
    pre[v]=_index++;
}

bool inQ[MAXV];
int dis[MAXV],Q[MAXV],prev[MAXV],curr[MAXV];

int SPFA(int src,int des)
{
    int u,v,w;
    memset(dis,inf,sizeof(dis));
    memset(inQ,false,sizeof(inQ)); dis[Q[0]=src]=0;

    for(int f=0,r=1;f!=r;f=(f+1)%MAXV)
    {
        inQ[u=Q[f]]=false;
        for(int i=pre[u];i!=-1;i=next[i])
        {
            v=G[i].v;
            w=G[i].c;
            if(G[i].w&&dis[v]>dis[u]+w)
            {
                dis[v]=dis[u]+w;
                prev[v]=u;
                curr[v]=i;

                if(!inQ[v])
                {
                    inQ[Q[r]=v]=true;
                    r=(r+1)%MAXV;
                }
            }
        }
    }

    return dis[des];
}
int MinCost(int src,int des,int flow)
{
    int cnt,sum=0,tot=0;
    while(SPFA(src,des)!=inf)
    {
        cnt=inf;
        for(int i=des;i!=src;i=prev[i])
        {
            cnt=min(cnt,G[curr[i]].w);
        }
        for(int i=des;i!=src;i=prev[i])
        {
            G[curr[i]].w-=cnt;
            G[curr[i]^1].w+=cnt;
        }
        sum+=(cnt*dis[des]);
        tot+=cnt;
    }
    return tot==flow?sum:-1;
}

int n,m,u,v,w;

int main()
{
    #ifndef ONLINE_JUDGE
    freopen("Cyclic Tour.txt","r",stdin);
    #endif

    while(scanf("%d %d",&n,&m)!=EOF)
    {
        clear();
        for(int i=0;m>i;i++)
        {
            scanf("%d %d %d",&u,&v,&w);
            add(u,v+n,1,w);
        }
        for(int i=0;n>i;i++)
        {
            add(0,i+1,1,0);
            add(i+n+1,n+n+1,1,0);
        }
        printf("%d\n",MinCost(0,2*n+1,n));
    }
    return 0;
}