// 题目来源：POJ 3683// 题目大意：有n对新人要举行仪式，每对都有两个时间段可以选择，问是否可以所有新人的仪式时间不重叠// 解决方法：2-sat的一眼题，把仪式可以选择的两个时间段看成两个点，仪式时间重叠的即为矛盾关系// 特别注意：输出要按输入顺序！不要按照自己处理的顺序来。#include <cstdio>#define o 10000 #define _ 1000000void link( int, int ); // 保存原图必须边 void link2( int, int ); // 保存缩点图（转成反边）bool judge( int, int, int, int ); // 判断时间段是否重叠void tarjan( int ); // 强连通分量缩点void topsort( int ); // 拓扑排序void dfs( int ); // 递归染色void print( int, int ); // 转换格式输出int next[_*2], g[_*2], next2[_], g2[_];int h[o], h2[o], stack[o], dfn[o], low[o], a[o], b[o], c[o], code[o], id[o], done[o], cfl[o], color[o];int n, t, t2, top, tt, index, cnt;char st[10];bool ins[o];int main( ){ freopen( "3683.in", "r", stdin ); freopen( "3683.out", "w", stdout ); scanf( "%d", &n ); for( int i = 1; i <= n; i++ ) { scanf( "%s", st ); a[i] = ( st[0]-48 )*600 + ( st[1]-48 )*60; a[i] += ( st[3]-48 )*10 + st[4]-48; scanf( "%s", st ); b[i] = ( st[0]-48 )*600 + ( st[1]-48 )*60; b[i] += ( st[3]-48 )*10 + st[4]-48; scanf( "%d", &c[i] ); } for( int i = 1; i <= n; i++ ) for( int j = i + 1; j <= n; j++ ) { if( !judge( a[i], a[i]+c[i], a[j], a[j]+c[j] ) ) { link( i, j+n ); link( j, i+n ); } if( !judge( a[i], a[i]+c[i], b[j]-c[j], b[j] ) ) { link( i, j ); link( j+n, i+n ); } if( !judge( b[i]-c[i], b[i], a[j], a[j]+c[j] ) ) { link( i+n, j+n ); link( j, i ); } if( !judge( b[i]-c[i], b[i], b[j]-c[j], b[j] ) ) { link( i+n, j ); link( j+n, i ); } } for( int i = 1; i <= 2*n; i++ ) if( !dfn[i] ) tarjan( i ); for( int i = 1; i <= n; i++ ) if( code[i] == code[i+n] ) { printf( "NO" ); return 0; } printf( "YES\n" ); int j; for( int i = 1; i <= 2*n; i++ ) for( int k = h[i]; k; k = next[k] ) { j = g[k]; if( code[i] != code[j] ) { link2( code[j], code[i] ); id[ code[i] ]++; } } for( int i = 1; i <= cnt; i++ ) if( id[i] == 0 ) topsort( i ); for( int i = 1; i <= n; i++ ) { cfl[ code[i] ] = code[i+n]; cfl[ code[i+n] ] = code[i]; } for( int ii = 1; ii <= cnt; ii++ ) { int i = done[ii]; if( color[i] != 0 ) continue; color[i] = 1; dfs( cfl[i] ); } for( int i = 1; i <= n; i++ ) if( color[ code[i] ] == 1 ) print( a[i], a[i]+c[i] ); else print( b[i]-c[i], b[i] );}void link( int aa, int bb ){ next[++t] = h[aa]; h[aa] = t; g[t] = bb;}void link2( int aa, int bb ){ next2[++t2] = h2[aa]; h2[aa] = t2; g2[t2] = bb;}bool judge( int aa, int bb, int cc, int dd ){ if( bb <= cc || dd <= aa ) return 1; return 0;}void tarjan( int i ){ int j; dfn[i] = low[i] = ++index; stack[++top] = i; ins[i] = true; for( int k = h[i]; k; k = next[k] ) { j = g[k]; if( !dfn[j] ) { tarjan( j ); if( low[j] < low[i] ) low[i] = low[j]; } else if( ins[j] && dfn[j] < low[i] ) low[i] = dfn[j]; } if( dfn[i] == low[i] ) { cnt++; do { j = stack[top--]; code[j] = cnt; ins[j] = false; } while( i != j ); }}void topsort( int i ){ int j; id[i] = -1; done[++tt] = i; for( int k = h2[i]; k; k = next2[k] ) { j = g2[k]; id[j]--; if( id[j] == 0 ) topsort( j ); }}void dfs( int i ){ int j; color[i] = 2; for( int k = h2[i]; k; k = next2[k] ) { j = g2[k]; if( color[j] == 0 ) dfs( j ); }}void print( int aa, int bb ){ if( aa / 60 < 10 ) printf( "0" ); printf( "%d:", aa / 60 ); if( aa % 60 < 10 ) printf( "0" ); printf( "%d ", aa % 60 ); if( bb / 60 < 10 ) printf( "0" ); printf( "%d:", bb / 60 ); if( bb % 60 < 10 ) printf( "0" ); printf( "%d\n", bb % 60 );}
