uva 11167 网络流 【好题】

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <queue>
#include <vector>
#include <map>
#define pb push_back
#define mp make_pair
#define eps 1e-9
#define zero(x) (fabs(x)<eps)
#define pi acos(-1.0)
#define f1 first
#define f2 second
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define initial 1,n,1
#define CLR(x,y) memset(x,y,sizeof(x))

using namespace std;
typedef long long LL;
typedef pair <int, int> PII;
template<typename X> inline bool minimize(X&p,X q){if(p<=q)return 0;p=q;return 1;}
template<typename X> inline bool maximize(X&p,X q){if(p>=q)return 0;p=q;return 1;}
const int MAXN=330,inf=0x3f3f3f3f;
vector<pair<int,int> > anss[105];
int vv[50005][5];

int a[105][3];
int tt[210];
int n,mm;
struct ISAP
{   struct Edge
    {
        int from,to,cap,flow;
        Edge(){}
        Edge(int a,int b,int c,int d):from(a),to(b),cap(c),flow(d){}
    };
    int n,m,s,t;//结点数，边数(含反向弧)，源点，汇点
    vector<Edge> edges;//边表，edges[e]&edges[e^1]互为反向弧
    vector<int> G[MAXN];//邻接表，G[i][j]表示结点i的第j条边在e数组中的序号
    bool vis[MAXN];//BFS使用
    int d[MAXN];//从起点到i的距离
    int cur[MAXN];//当前弧下标
    int p[MAXN];//可增广路上的上一条弧
    int num[MAXN];//距离标号计数

    void AddEdge(int from,int to,int cap)//重边不影响
    {
        edges.push_back(Edge(from,to,cap,0));
        edges.push_back(Edge(to,from,0,0));//容量为0,表示反向弧
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    void init(int n)
    {
        this->n=n;
        for(int i=0;i<n;++i) G[i].clear();
        edges.clear();
    }

    void BFS()//反向
    {   for (int i=0;i<n;i++)
            d[i]=n+10; //!!!
        memset(vis,0,sizeof(vis));
        queue<int> Q;
        Q.push(t);
        d[t]=0;
        vis[t]=1;
        while(!Q.empty())
        {
            int x=Q.front();
            Q.pop();
            for(int i=0; i<G[x].size(); ++i)
            {
                Edge& e=edges[G[x][i]^1];
                if(!vis[e.from]&&e.cap>e.flow)
                {
                    vis[e.from]=1;
                    d[e.from]=d[x]+1;
                    Q.push(e.from);
                }
            }
        }
    }

    int Augment()
    {
        int x=t,a=inf;
        while(x!=s)
        {
            Edge& e=edges[p[x]];
            a=min(a,e.cap-e.flow);
            x=edges[p[x]].from;
        }
        x=t;
        while(x!=s)
        {
            edges[p[x]].flow+=a;
            edges[p[x]^1].flow-=a;
            x=edges[p[x]].from;
        }
        return a;
    }

    int Maxflow(int s,int t)//结点数
    {
        this->s=s,this->t=t;
        int flow=0;
        BFS();
        memset(num,0,sizeof(num));
        for(int i=0;i<n;++i) if (d[i]!=n+10)++num[d[i]];//!!!
        int x=s;
        memset(cur,0,sizeof(cur));
        while(d[s]<n)
        {
            if(x==t)
            {
                flow+=Augment();
                x=s;
            }
            int ok=0;
            for(int i=cur[x];i<G[x].size();++i)
            {
                Edge& e=edges[G[x][i]];
                if(e.cap>e.flow&&d[x]==d[e.to]+1)//Advance
                {
                    ok=1;
                    p[e.to]=G[x][i];
                    cur[x]=i;
                    x=e.to;
                    break;
                }
            }
            if(!ok)//Retreat
            {
                int m=n-1;
                for(int i=0;i<G[x].size();++i)
                {
                    Edge& e=edges[G[x][i]];
                    if(e.cap>e.flow) m=min(m,d[e.to]);
                }
                if(--num[d[x]]==0) break;//gap优化
                num[d[x]=m+1]++;
                cur[x]=0;
                if(x!=s) x=edges[p[x]].from;
            }
        }
        return flow;
    }
    void solve(int nn)
    {   memset(vv,0,sizeof(vv));
        int tr=-1,tl=-1,last=-1;
        for (int i=0;i<edges.size();i++)
        if (edges[i].from<=nn&&edges[i].from>=1&&edges[i].flow>0)
                 {   int index=edges[i].to-nn-1,res=edges[i].flow;
                     for(int k=0;k<mm;k++)
                        {
                            if(vv[index][k]==tt[index+1]-tt[index])continue;
                            if(vv[index][k]+res<=tt[index+1]-tt[index])
                                {
                                    anss[edges[i].from].pb(mp(tt[index]+vv[index][k],tt[index]+vv[index][k]+res));
                                    vv[index][k]+=res;
                                    break;
                                }
                            else
                                {
                                    res-=(tt[index+1]-tt[index]-vv[index][k]);
                                    anss[edges[i].from].pb(mp(tt[index]+vv[index][k],tt[index+1]));
                                    vv[index][k]=tt[index+1]-tt[index];
                                    if(!res)break;
                                }
                        }
                 }

    }
}it;

void doit()
{   scanf("%d",&mm);
    int id=-1,sum=0;
    for (int i=1;i<=n;i++)
        {scanf("%d%d%d",&a[i][0],&a[i][1],&a[i][2]);
         tt[++id]=a[i][1];
         tt[++id]=a[i][2];
         sum+=a[i][0];
        }
    sort(tt,tt+id+1);
    int cnt=unique(tt,tt+id+1)-tt;


    it.init(n+cnt+1);
    int s=0,t=n+cnt;

    for (int i=1;i<=n;i++)
    for (int j=1;j<cnt;j++)
    if (a[i][1]<=tt[j-1]&&a[i][2]>=tt[j])
        it.AddEdge(i,n+j,tt[j]-tt[j-1]);
    for (int i=1;i<=n;i++)  it.AddEdge(0,i,a[i][0]);
    for (int i=1;i<cnt;i++) it.AddEdge(n+i,t,(tt[i]-tt[i-1])*mm);

    int ans=it.Maxflow(s,t);
     for (int i=1;i<=n;i++)anss[i].clear();
    if (ans!=sum)
         {   printf("No\n");
             return;
         }
    printf("Yes\n");
    it.solve(n);

    for (int i=1;i<=n;i++)
        {   sort(anss[i].begin(),anss[i].end());
            for(int j=1;j<anss[i].size();j++)
            if(anss[i][j].f1==anss[i][j-1].f2)
                {
                    anss[i][j-1].f2=anss[i][j].f2;
                    anss[i].erase(anss[i].begin()+j);
                    j--;
                }

            printf("%d",anss[i].size());
            for (int j=0;j<anss[i].size();j++)
                printf(" (%d,%d)",anss[i][j].f1,anss[i][j].f2);
            printf("\n");
        }
}
int main()
{   //freopen("in.txt","r",stdin);
    int id=0;
    while (scanf("%d",&n),n)
        {printf("Case %d: ",++id);
         doit();
        }
}
/*
3 2
2 1 4
2 1 4
2 1 4

*/