poj3683

Priest John's Busiest Day

Time Limit: 2000MS		Memory Limit: 65536K

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
题意：题意：一个牧师必须参加所给出所有婚礼，可以在Si参加到Si+Di或者从Ti-Di参加到Ti，参加1个即可，牧师不能
同时参加两个婚礼，求能否参加并且输出参加时间表。
题解：典型的2-sat题目，把从一开始参加的时间段和从中途参加的时间段看成两个点，如果两个点在同一强连通分量中，就不行.

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<string>
#include<queue>
using namespace std;
// 0--n-1 1   n -- 2n-1 0
const int maxn = 2005, maxm = 5000000;
int tot,h[maxn], hh[maxn],ru[maxn],t[maxn],top,dfn[maxn],low[maxn],idx,place[maxn],scccnt;
int opp[maxn], color[maxn],st[maxn], ed[maxn], st1[maxn], ed1[maxn], d[maxn];
int n,m;
queue <int> Q;vector <int> vec[maxn];
bool ins[maxn];
struct edge{
    int u,to,nxt;
}G[maxm];
int get(string a){
    int x = 60*(a[0]-'0')*10+60*(a[1]-'0');
    int y = (a[3]-'0')*10+a[4]-'0';
    return x+y;
}

void add(int u,int v){G[++tot].nxt = h[u];G[tot].u = u;G[tot].to = v;h[u] = tot;}
bool check(int i, int j){
    if(st[i] <= ed1[j] && ed1[j] < st1[i] && st1[i] <= ed[j])return 1;
    if(st[i] >= ed1[j] && ed[j] > st[i] && st1[i] >= ed[j])return 1;
    if(st[i] <= ed1[j] && ed[j] <= st1[i])return 1;
    if(st[i] >= ed1[j] && ed[j] >= st1[i])return 1;
    return 0;
}
void build(int i, int j){
    if(st[i] > st[j])swap(i, j);
    if(st[j]<st1[i]){
        add(i, j+n);
        add(j, i+n);
    }
    if(ed1[i] > ed1[j])swap(i, j);
    if(ed1[j]<ed[i]){
        add(j+n, i);
        add(i+n, j);
    }
    if(check(i, j)){//sti edj
        add(i, j);
        add(j+n, i+n);
    }
    if(check(j, i)){//stj edi
        add(j, i);
        add(i+n, j+n);
    }
} 
void tarjan(int x){
    dfn[x] = low[x] = ++idx;
    t[++top] = x, ins[x] = 1;
    for(int i = h[x]; i; i = G[i].nxt){
        int v = G[i].to;
        if(!dfn[v]){
            tarjan(v);
            low[x] = min(low[x], low[v]);
        }
        else if(ins[v]) low[x] = min(low[x], dfn[v]);
        
    }
    
    if(dfn[x] == low[x]){
        scccnt++;
        while(1){
            place[t[top]] = scccnt;
            ins[t[top--]] = 0;
            if(t[top+1] == x)break;
        }
    }
}
void topsort(){
    for(int i = 1; i <= scccnt; i++)
        if(!ru[i])Q.push(i);
    while(!Q.empty()){
        int u = Q.front(); Q.pop();
        if(!color[u]){
            color[u] = 1;
            color[opp[u]] = 2;
        }
        for(int i = 0; i < vec[u].size(); i++){
            int v = vec[u][i];
            ru[v]--;
            if(!ru[v]) Q.push(v);
        }        
    }
    
    
}
void tb(int i){
    int qs = st[i], qd = st1[i];
    if(i > n)qs = ed1[i-n], qd = ed[i-n];
    printf("%02d:%02d %02d:%02d
",qs/60,qs%60,qd/60,qd%60);
    
}
int main(){
    
    bool ans = 1;
    scanf("%d",&n);
    for(int i = 1; i <= n; i++){
        string a, b;
        cin>>a>>b;
        scanf("%d",&d[i]);
        st[i] = get(a);
        st1[i] = st[i]+d[i];
        ed[i] = get(b);
        ed1[i] = ed[i]-d[i];
    }
    for(int i = 1; i <= n; i++)
        for(int j = i+1; j <= n; j++)
            build(i, j);
    for(int i = 1; i <= 2*n; i++)
        if(!dfn[i])tarjan(i);
    for(int i = 1; i <= n; i++)
        if(place[i] == place[i+n])ans = 0;
        else opp[place[i]] = place[i+n], opp[place[i+n]] = place[i];
    if(!ans){
        puts("NO");return 0; 
    }
    //for(int i = 0; i<n ;i++)cout<<place[i]<<" "<<place[i+n]<<endl;
    for(int i = 1; i <= 2*n; i++)
        for(int j = h[i]; j; j = G[j].nxt){
            int v = G[j].to;
            if(place[v] != place[i])
                vec[place[v]].push_back(place[i]), ru[place[i]]++;
        }
    topsort();
    puts("YES");
    for(int i = 1; i <=n; i++)
        if(color[place[i]]==1)tb(i);
        else tb(i+n);
    
}
/*
5
10:00 10:20 10
11:00 12:00 30
11:30 12:40 20
10:15 11:00 20
10:30 11:00 25
*/