POJ2175-最小费用流消圈算法

#include <cstdio>
#include <queue>
#include <stack>
#include <cstring>
#include <iostream>
using namespace std;

const int NN=210;
const int MM=100000;
const int INF=0x3fffffff;

inline int Abs(int x)
{
    if (x>=0) return x;
    else      return -x;
}

int n,m,S,T,NV,en,head[NN];
struct Edge
{
    int u,v,f,c,next;
} e[MM];
inline void add(int u,int v,int c,int f1,int f2)
//这题的很多版本的流量都是用的邻接矩阵,我写惯了邻接表,强迫症下不喜欢矩阵了~
{
    e[en].u=u;
    e[en].v=v;
    e[en].c=c;
    e[en].f=f1;
    e[en].next=head[u];
    head[u]=en++;
    e[en].u=v;
    e[en].v=u;
    e[en].c=-c;
    e[en].f=f2;
    e[en].next=head[v];
    head[v]=en++;
}

int dis[NN],cou[NN],p[NN];
bool vis[NN];
int circle_spfa()
{
    int i,u,v;

    stack<int> q;//找圈的spfa用栈感觉比用队列好,栈是深度优先的
    for (i=0; i<NV; i++) dis[i]=INF,cou[i]=vis[i]=0;
    dis[T]=0; cou[T]++;
    q.push(T);

    while (!q.empty())
    {
        u=q.top(); q.pop();
        vis[u]=false;
        for (i=head[u]; i!=-1; i=e[i].next)
        {
            v=e[i].v;
            if (e[i].f && dis[v]>dis[u]+e[i].c)
            {
                dis[v]=dis[u]+e[i].c;
                p[v]=i;
                if (!vis[v])
                {
                    q.push(v);
                    vis[v]=true;
                }
                if (++cou[v]>=NV) return v;
            }
        }
    }
    return -1;
}

int ans[NN][NN];
void solve()
{
    int i,j,x,u,v;

    x=circle_spfa();//这题不需要求最优,spfa找负圈找一次即可

    if (x==-1) { puts("OPTIMAL"); return; }
    puts("SUBOPTIMAL");  //单词拼错，贡献WA n次

    memset(vis,0,sizeof(vis));
    while (!vis[x])
    {
        vis[x]=true;
        x=e[p[x]].u;
    }//spfa退出的点未必是负圈中的点

    v=x; do
    {
        u=e[p[v]].u;
        e[p[v]].f--;
        e[p[v]^1].f++;
        v=u;
    }while (v!=x);//消圈

    memset(ans,0,sizeof(ans));
    for (i=0; i<en; i+=2)
    {
        if (!e[i^1].f || e[i].u==S || e[i].v==T) continue;
        ans[e[i].u][e[i].v-n]=e[i^1].f;
    }
    for (i=1; i<=n; i++)
    {
        for (j=1; j<m; j++) printf("%d ",ans[i][j]);
        printf("%d\n",ans[i][m]);
    }
}

int x[NN],y[NN],z[NN],s1[NN],s2[NN];
int main()
{
    int i,j,diss,w;

    while (~scanf("%d%d",&n,&m))
    {
        en=S=0; T=n+m+1; NV=T+1;
        memset(head,-1,sizeof(head));

        for (i=1; i<=n+m; i++) scanf("%d%d%d",&x[i],&y[i],&z[i]);
        for (i=1; i<=m; i++) s2[i]=0;
        for (i=1; i<=n; i++)
        {
            s1[i]=0;
            for (j=1; j<=m; j++)
            {
                scanf("%d",&w);
                diss=Abs(x[i]-x[j+n])+Abs(y[i]-y[j+n])+1;
                s1[i]+=w;
                s2[j]+=w;
                add(i,j+n,diss,INF-w,w);
            }
        }
        for (i=1; i<=n; i++) add(S,i,0,z[i]-s1[i],s1[i]);
        for (i=1; i<=m; i++) add(i+n,T,0,z[i+n]-s2[i],s2[i]);
        solve();
    }
    return 0;
}