HDU 4494 Teamwork 2013通化邀请赛 B题 费用流

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <math.h>

using namespace std;
const int maxn=500;
const int inf=1<<30;

double dist[maxn][maxn];
int x[maxn],y[maxn],star[maxn],p[maxn],a[maxn][10];
struct edge
{
    int u,v,c,w,next;
}e[maxn*maxn];
int tot;
int head[maxn];
int from[maxn];
bool vis[maxn];
int queue[maxn*maxn],d[maxn];
int n,m;
double dis(int i,int j)
{
    return sqrt(((double)x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
}
void add(int u,int v,int c,int w)//加边，u->v,流量c，花费w;
{
    tot++;
    e[tot].u=u;
    e[tot].v=v;
    e[tot].c=c;
    e[tot].w=w;
    e[tot].next=head[u];
    head[u]=tot;
}
bool spfa(int star,int end)
{
    for (int i=star;i<=end;i++)
        d[i]=inf;
    memset(vis,0,sizeof(vis));
    memset(from,-1,sizeof(from));//到达该点的流从哪条边流进的；
    int h=0,t=1;//队列的头尾指针；
    d[star]=0;//到达该点的最小花费；
    vis[star]=1;//是否入队的标记；
    queue[1]=star;//队列；
    while(h<t)
    {
        h++;
        int u=queue[h];
        for(int i=head[u];i!=-1;i=e[i].next)
        if(e[i].c>0)
        {
            int v=e[i].v,w=e[i].w;
            if(d[v]>d[u]+w)
            {
                d[v]=d[u]+w;
                from[v]=i;
                if(!vis[v])
                {
                    vis[v]=1;
                    t++;
                    queue[t]=v;
                }
            }
        }
        vis[u]=0;
    }
    return from[end]!=-1;
}
int maxflow(int s,int t)//网络流的s点，t点的编号；
{
    int ans=0;
    int flow=0;
    while(spfa(s,t))
    {
        for(int i=from[t];i!=-1;i=from[e[i].u])//倒推找到的流量最小的流；
        {
            e[i].c--;
            e[i^1].c++;
            ans+=e[i].w;
        } 
        flow++;
    }
    return ans;//返回值是最大流的最小费用；
   //return flow;
}
int main()
{
   
    int t;
    int x0,y0;
    scanf("%d",&t);
    while (t--)
    {
        scanf("%d%d",&n,&m);
        scanf("%d%d",&x0,&y0);
        for (int i=1;i<n;i++)
        {
            scanf("%d%d%d%d",&x[i],&y[i],&star[i],&p[i]);
            for (int j=1;j<=m;j++)
            scanf("%d",&a[i][j]);
        }

        for (int i=1;i<n;i++)
        for (int j=i+1;j<n;j++)
        dist[i][j]=dist[j][i]=dis(i,j);

        int ans=0;
        n--;
        for (int j=1;j<=m;j++)
        {
            tot=-1;
            memset(head,-1,sizeof(head));
            for (int i=1;i<=n;i++)
            {
                add(0,i,a[i][j],1);
                add(i,0,0,-1);

                add(i,i+n*2,a[i][j],0);
                add(i+n*2,i,0,0);

                add(i+n*2,3*n+1,a[i][j],0);
                add(3*n+1,i+n*2,0,0);

                add(0,i+n,a[i][j],0);
                add(i+n,0,0,0);
            }

            for (int i=1;i<=n;i++)
            for (int k=1;k<=n;k++)
            if (i!=k)
            {
                if (star[i]+p[i]+dist[i][k]<=star[k])
                {
                    add(i+n,k+2*n,min(a[i][j],a[k][j]),0);
                    add(k+2*n,i+n,0,0);
                }
            }

            ans+=maxflow(0,3*n+1);
       }
            printf("%d\n",ans);
    }
    return 0;
}

看来学长的另一种做法，觉得也不错，给大家推荐一下http://blog.youkuaiyun.com/yrleep/article/details/13391427