URAL 1774 A - Barber of the Army of Mages 最大流

[b]A - Barber of the Army of Mages[/b]
Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接
http://acm.hust.edu.cn/vjudge/contest/view.action?cid=87643#problem/A
Description

Petr, elected as a warlord of the army of mages, faced a challenging problem. All magicians recruited in the army had heavy beards, which were quite unacceptable for soldiers. Therefore, Petr ordered all recruits to shave their beards as soon as possible. Of course, all magicians refused to do it, referring to the fact they don't know any shaving spell. Fortunately, a magician Barberian agreed to shave all recruits.

Barberian can cast a “Fusion Power” spell which shaves beards of at most k magicians in one minute. In order to achieve full effect every magician should be shaved twice: the first spell shaves close, the second spell shaves even closer. For each recruit Petr appointed a time when he should visit Barberian. Unfortunately, the discipline in the new army is still far from perfect, so every magician will come to Barberian in time, but everyone will wait for the shave until his patience is exhausted and will disappear after that.

Determine whether Barberian will be able to shave beards of all magicians before they disappear.

Input

The first line contains two space-separated integers n and k (1 ≤ n, k ≤ 100) , which are the number of recruits in the army and the number of magicians Barber can shave simultaneously. The i-th of the following n lines contains space-separated integers ti and si(0 ≤ ti ≤ 1000; 2 ≤ si ≤ 1000) , which are the time in minutes, at which the i-th magician must come to Barberian, and the time in minutes he is ready to spend there, including shaving time.

Output

If Barberian is able to shave beards of all magicians, output “Yes” in the first line. The i-th of the following n lines should contain a pair of integers pi, qi, which are the moments at which Barberian should cast the spell on the i-th magician ( ti ≤ pi < qi ≤ ti + si − 1) . If at least one magician disappears before being completely shaved, output a single word “No”.

Sample Input

3 2
1 3
1 3
1 3

Sample Output

Yes
1 2
1 3
2 3

HINT

题意

有n个顾客，每个顾客需要理2次胡须

每一秒，理发师可以给k个人理发

然后每个顾客必须在x秒到x+y-1秒内理完

然后让你构造出一种解

题解：

网络流，贪心的话就走远了……

S-2-顾客-1-天数-k-T

然后跑一发最大流就好了

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 5000
#define mod 10007
#define eps 1e-9
int Num;
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
inline ll read()
{
ll x=0,f=1;char ch=getchar();
while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}
//**************************************************************************************

namespace NetFlow
{
const int MAXN=100000,MAXM=500000,inf=1e9;
vector<int> Q[1100];
struct Edge
{
int v,c,f,nx;
Edge() {}
Edge(int v,int c,int f,int nx):v(v),c(c),f(f),nx(nx) {}
} E[MAXM];
int G[MAXN],cur[MAXN],pre[MAXN],dis[MAXN],gap[MAXN],N,sz;
void init(int _n)
{
N=_n,sz=0; memset(G,-1,sizeof(G[0])*N);
}
void link(int u,int v,int c)
{
E[sz]=Edge(v,c,0,G[u]); G[u]=sz++;
E[sz]=Edge(u,0,0,G[v]); G[v]=sz++;
}
int ISAP(int S,int T)
{//S -> T
int maxflow=0,aug=inf,flag=false,u,v;
for (int i=0;i<N;++i)cur[i]=G[i],gap[i]=dis[i]=0;
for (gap[S]=N,u=pre[S]=S;dis[S]<N;flag=false)
{
for (int &it=cur[u];~it;it=E[it].nx)
{
if (E[it].c>E[it].f&&dis[u]==dis[v=E[it].v]+1)
{
if (aug>E[it].c-E[it].f) aug=E[it].c-E[it].f;
pre[v]=u,u=v; flag=true;
if (u==T)
{
for (maxflow+=aug;u!=S;)
{
E[cur[u=pre[u]]].f+=aug;
E[cur[u]^1].f-=aug;
}
aug=inf;
}
break;
}
}
if (flag) continue;
int mx=N;
for (int it=G[u];~it;it=E[it].nx)
{
if (E[it].c>E[it].f&&dis[E[it].v]<mx)
{
mx=dis[E[it].v]; cur[u]=it;
}
}
if ((--gap[dis[u]])==0) break;
++gap[dis[u]=mx+1]; u=pre[u];
}
return maxflow;
}
bool bfs(int S,int T)
{
static int Q[MAXN]; memset(dis,-1,sizeof(dis[0])*N);
dis[S]=0; Q[0]=S;
for (int h=0,t=1,u,v,it;h<t;++h)
{
for (u=Q[h],it=G[u];~it;it=E[it].nx)
{
if (dis[v=E[it].v]==-1&&E[it].c>E[it].f)
{
dis[v]=dis[u]+1; Q[t++]=v;
}
}
}
return dis[T]!=-1;
}
int dfs(int u,int T,int low)
{
if (u==T) return low;
int ret=0,tmp,v;
for (int &it=cur[u];~it&&ret<low;it=E[it].nx)
{
if (dis[v=E[it].v]==dis[u]+1&&E[it].c>E[it].f)
{
if (tmp=dfs(v,T,min(low-ret,E[it].c-E[it].f)))
{
ret+=tmp; E[it].f+=tmp; E[it^1].f-=tmp;
}
}
}
if (!ret) dis[u]=-1; return ret;
}
int dinic(int S,int T)
{
int maxflow=0,tmp;
while (bfs(S,T))
{
memcpy(cur,G,sizeof(G[0])*N);
while (tmp=dfs(S,T,inf)) maxflow+=tmp;
}
return maxflow;
}
void solve(int S,int T,int p,int n)
{
int C = dinic(S,T);
if(C!=p)
printf("No\n");
else
{
printf("Yes\n");
for(int i=1;i<=n;i++)
{
int flag=0;
for(int j=G[i+10];j!=-1;j=E[j].nx)
{
if(E[j].f==1)
{
Q[i].push_back(E[j].v-150);
}
}
printf("%d %d\n",Q[i][1],Q[i][0]);
}
}
}
}

using namespace NetFlow;
int vis[5200];
int main()
{
init(5000);
int n=read(),k=read();
for(int i=11;i<=10+n;i++)
link(1,i,2);
for(int i=1;i<=n;i++)
{
int x=read(),y=read();
for(int j=x;j<=x+y-1;j++)
{
link(i+10,150+j,1);
if(!vis[j])
{
vis[j]=1;
link(150+j,3000,k);
}
}
}
solve(1,3000,2*n,n);
//cout<<dinic(1,4000)<<endl;
}