图论（2-sat）：Priest John's Busiest Day

Priest John's Busiest Day

 

Description

John is the only priest in his town. September 1st is the John's busiest day in a year because there is an old legend in the town that the couple who get married on that day will be forever blessed by the God of Love. This year N couples plan to get married on the blessed day. The i-th couple plan to hold their wedding from time S_i to time T_i. According to the traditions in the town, there must be a special ceremony on which the couple stand before the priest and accept blessings. The i-th couple need D_i minutes to finish this ceremony. Moreover, this ceremony must be either at the beginning or the ending of the wedding (i.e. it must be either from S_i to S_i + D_i, or from T_i - D_i to T_i). Could you tell John how to arrange his schedule so that he can present at every special ceremonies of the weddings.

Note that John can not be present at two weddings simultaneously.

Input

The first line contains a integer N ( 1 ≤ N ≤ 1000).
The next N lines contain the S_i, T_i and D_i. S_i and T_i are in the format of hh:mm.

Output

The first line of output contains "YES" or "NO" indicating whether John can be present at every special ceremony. If it is "YES", output another N lines describing the staring time and finishing time of all the ceremonies.

Sample Input

2
08:00 09:00 30
08:15 09:00 20

Sample Output

YES
08:00 08:30
08:40 09:00

　　闹半天结果是数组开小了。

　　这里是2-sat O（n+m）输出任意解的模板。

#include <iostream>
#include <cstring>
#include <cstdio>
using namespace std;
const int maxn=4010;
const int maxm=4000010;
int n;
struct E{
    int cnt,fir[maxn],to[maxm],nxt[maxm];
    void addedge(int a,int b){
        nxt[++cnt]=fir[a];
        fir[a]=cnt;
        to[cnt]=b;
    }
}e,g;

struct Time{
    int h,s;
    Time(int h_=0,int s_=0){
        h=h_;s=s_;
    }
    friend bool operator <=(Time x,Time y){
        return x.h!=y.h?x.h<y.h:x.s<=y.s;
    }
    friend Time operator +(Time x,Time y){
        Time ret(0,0);
        ret.s=(x.s+y.s)%60;
        ret.h=x.h+y.h+(x.s+y.s)/60;
        return ret;
    }
    friend Time operator -(Time x,Time y){
        Time ret(0,0);
        ret.s=x.s-y.s+(x.s<y.s?1:0)*60;
        ret.h=x.h-y.h-(x.s<y.s?1:0);
        return ret;
    }
    void Print(){
        printf("%02d:%02d",h,s);
    }
}p1[maxn],p2[maxn],p3[maxn],p4[maxn];

bool Judge(Time pl,Time pr,Time ql,Time qr){
    if(pr<=ql||qr<=pl)return false;
    return true;
}

int tot,ID[maxn],low[maxn],scnt;
int st[maxn],top,scc[maxn];

void Tarjan(int x){
    ID[x]=low[x]=++tot;st[++top]=x;
    for(int i=e.fir[x];i;i=e.nxt[i])
        if(!ID[e.to[i]]){
            Tarjan(e.to[i]);
            low[x]=min(low[x],low[e.to[i]]);
        }
        else if(!scc[e.to[i]])
            low[x]=min(low[x],ID[e.to[i]]);
    if(low[x]==ID[x]){
        ++scnt;
        while(true){  
            int y=st[top--];
            scc[y]=scnt;
            if(x==y)break;
        }
    }
}

bool Solve(){
    for(int i=0;i<n*2;i++)
        if(!ID[i])Tarjan(i);
    
    for(int i=0;i<n;i++)
        if(scc[i*2]==scc[i*2+1])
            return false;
    return true;        
}

int rev[maxn];
int match[maxn],in[maxn];
int q[maxn],front,back;
void Topo(){
    for(int i=0;i<n;i++){
        rev[scc[i*2]]=scc[i*2+1];
        rev[scc[i*2+1]]=scc[i*2];
    }
    for(int x=0;x<n*2;x++)
        for(int i=e.fir[x];i;i=e.nxt[i])
            if(scc[x]!=scc[e.to[i]])
                g.addedge(scc[e.to[i]],scc[x]),in[scc[x]]+=1;
                
    for(int i=1;i<=scnt;i++)
        if(!in[i])q[back++]=i;
    
    while(front<back){
        int x=q[front++];
        if(match[x]==0){
            match[x]=1;
            match[rev[x]]=2;
        }
        for(int i=g.fir[x];i;i=g.nxt[i])
            if(--in[g.to[i]]==0)q[back++]=g.to[i];
    }    
}

void Print(){
    for(int i=0;i<n;i++){
        if(match[scc[i*2]]==1){
            p1[i].Print();printf(" ");
            p2[i].Print();printf("
");
        }
        else{
            p3[i].Print();printf(" ");
            p4[i].Print();printf("
");
        }
    }
}

int main(){
    scanf("%d",&n);
    for(int i=0,d;i<n;i++){
        scanf("%d:%d",&p1[i].h,&p1[i].s);
        scanf("%d:%d",&p4[i].h,&p4[i].s);
        scanf("%d",&d);
        p2[i]=p1[i]+Time(d/60,d%60);
        p3[i]=p4[i]-Time(d/60,d%60);
    }
    
    for(int i=0;i<n;i++)
        for(int j=i+1;j<n;j++){
            if(Judge(p1[i],p2[i],p1[j],p2[j])){
                e.addedge(i*2,j*2+1);
                e.addedge(j*2,i*2+1);
            }
            if(Judge(p1[i],p2[i],p3[j],p4[j])){
                e.addedge(i*2,j*2);
                e.addedge(j*2+1,i*2+1);
            }
            if(Judge(p3[i],p4[i],p1[j],p2[j])){
                e.addedge(i*2+1,j*2+1);
                e.addedge(j*2,i*2);
            }
            if(Judge(p3[i],p4[i],p3[j],p4[j])){
                e.addedge(i*2+1,j*2);
                e.addedge(j*2+1,i*2);
            }
        }
    
    if(!Solve())
        printf("NO
");
    else{
        printf("YES
");
        Topo();Print();
    }
    return 0;
}