【题解】Priest John's Busiest Day POJ - 3683 ⭐⭐⭐ 【2-SAT 拓扑序】

Priest John’s Busiest Day POJ - 3683

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 Si to time Ti. 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 Di 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 Si to Si + Di, or from Ti - Di to Ti). 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 Si, Ti and Di. Si and Ti 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.

Examples

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

Sample Output
YES
08:00 08:30
08:40 09:00

Hint

题意:

有n个婚礼，每个婚礼有起始时间si，结束时间ti，还有一个主持时间ti，ti必须安排在婚礼的开始或者结束，
主持由祭祀来做，但是只有一个祭祀，所以各个婚礼的主持时间不能重复，问你有没有可能正常的安排主持时间，

不能输出no，能的话要输出具体的答案：即每个婚礼的主持时间段是什么样的。

题解:

转化为2-SAT问题来考虑, 每场婚礼分为开始和结束来考虑, 找出所有可能的冲突

对于婚礼i和婚礼j。i表示在开始主持，i2表示在结束主持，j类似。

枚举每一对不同的i和j。

如果i和j冲突。连接i j2

如果i和j2冲突，连接i j

如果i2和j冲突，连接i2 j2

如果i2和j2冲突，连接i2 j
接下来遍历1-N, 缩点后强连通分量拓扑序在前则开始时举行, 否则结束时举行, 有冲突则无法举行

经验小结:

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <string>
#include <stdlib.h>
#include <vector>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
using namespace std;
#define ms(x, n) memset(x,n,sizeof(x));
typedef  long long LL;
const int inf = 1<<30;
const LL maxn = 1010*4;int N, M, ecnt, head[maxn];
struct node{int to, next;
}es[maxn*maxn];
void addEdge(int u, int v){es[ecnt].to = v, es[ecnt].next = head[u];head[u] = ecnt++;
}
int indx = 0, scc = 0;
bool ins[maxn];
int dfn[maxn], low[maxn], belong[maxn];
stack<int> stk;
void Tarjan(int u){int v;dfn[u] = low[u] = ++indx;stk.push(u);ins[u] = true;for(int i = head[u]; i != -1; i = es[i].next){int v = es[i].to;if(!dfn[v]){Tarjan(v);low[u] = min(low[u], low[v]);}else if(ins[v]){low[u] = min(low[u], dfn[v]);}}if(dfn[u] == low[u]){++scc;do{v = stk.top();stk.pop();ins[v] = false;belong[v] = scc;}while(u != v);}
}
bool solve(){for(int i = 0; i < N*2; ++i)if(!dfn[i])Tarjan(i);for(int i = 0; i < N; ++i)if(belong[i] == belong[i+N])return false;return true;
}
int s[maxn], e[maxn], d[maxn];
int main()
{ms(head, -1);int h1, m1, h2, m2;scanf("%d",&N);for(int i = 0; i < N; ++i){scanf("%d:%d %d:%d %d",&h1,&m1,&h2,&m2,&d[i]);s[i] = h1*60+m1, e[i] = h2*60+m2;}//i表示i在开始时, i+N表示i在结束时for(int i = 0; i < N; ++i){for(int j = 0; j < i; ++j){//i在开始时和j在开始时冲突if(min(s[i]+d[i], s[j]+d[j]) > max(s[i], s[j])){addEdge(i, N+j);addEdge(j, N+i);}//i在结束时和j在结束时冲突if(max(e[i]-d[i], e[j]-d[j]) < min(e[i], e[j])){addEdge(i+N, j);addEdge(j+N, i);}//i在结束时和j在开始时冲突if(max(e[i]-d[i], s[j]) < min(e[i], s[j]+d[j])){addEdge(i+N, j+N);addEdge(j, i);}//i在开始时和j在结束时冲突if(max(e[j]-d[j], s[i]) < min(e[j], s[i]+d[i])){addEdge(j+N, i+N);addEdge(i, j);}}}if(solve()){printf("YES\n");for(int i = 0; i < N; ++i){if(belong[i] < belong[N+i]) //开始的拓扑序在前, 则在开始时举行典礼printf("%02d:%02d %02d:%02d\n",s[i]/60,s[i]%60,(s[i]+d[i])/60,(s[i]+d[i])%60);elseprintf("%02d:%02d %02d:%02d\n",(e[i]-d[i])/60,(e[i]-d[i])%60,e[i]/60,e[i]%60);}}else printf("NO\n");return 0;
}