题目分析:本题比较简单,将队长作为A,两位队员一组都作为A',这样就能顺利的建图了。
用idx[ ]表示所有队员调整过的编号,如第 i 队的队长a,idx[ a ] = i << 1,两位队员b,c编号idx[ b ] = idx[ c ] = i << 1 | 1
如果存在矛盾关系A,B,则建边< idx[ A ] , idx[ B ] ' > , < idx[ B ] , idx[ A ] ' >。(idx[ A ] ' = idx[ A ] ^ 1)
最后强连通判断是否有解即可。
代码如下:
#include <cstdio> #include <cstring> #include <algorithm> using namespace std ; #define REP( i , a , b ) for ( int i = ( a ) ; i < ( b ) ; ++ i ) #define FOR( i , a , b ) for ( int i = ( a ) ; i <= ( b ) ; ++ i ) #define REV( i , a , b ) for ( int i = ( a ) ; i >= ( b ) ; -- i ) #define CLR( a , x ) memset ( a , x , sizeof a ) const int MAXN = 2005 ; const int MAXM = 3005 ; const int MAXE = 23333 ; struct Edge { int v ; Edge* next ; } ; Edge E[MAXE] , *H[MAXN] , *cur ; int dfn[MAXN] , low[MAXN] , scc[MAXN] ; int S[MAXN] , top ; int dfs_clock , scc_cnt ; int n , m ; int idx[MAXM] ; void init () { top = dfs_clock = scc_cnt = 0 ; cur = E ; CLR ( H , 0 ) ; CLR ( scc , 0 ) ; CLR ( dfn , 0 ) ; } void addedge ( int u , int v ) { cur -> v = v ; cur -> next = H[u] ; H[u] = cur ++ ; } void tarjan ( int u ) { dfn[u] = low[u] = ++ dfs_clock ; S[top ++] = u ; for ( Edge* e = H[u] ; e ; e = e -> next ) { int v = e -> v ; if ( !dfn[v] ) { tarjan ( v ) ; low[u] = min ( low[u] , low[v] ) ; } else if ( !scc[v] ) low[u] = min ( low[u] , dfn[v] ) ; } if ( dfn[u] == low[u] ) { ++ scc_cnt ; while ( 1 ) { int v = S[-- top] ; scc[v] = scc_cnt ; if ( v == u ) break ; } } } void scanf ( int& x , char c = 0 ) { while ( ( c = getchar () ) < '0' || c > '9' ) ; x = c - '0' ; while ( ( c = getchar () ) >= '0' && c <= '9' ) x = x * 10 + c - '0' ; } int solve () { int a , b , c ; init () ; REP ( i , 0 , n ) { scanf ( a ) , scanf ( b ) , scanf ( c ) ; idx[a] = i << 1 ; idx[b] = idx[c] = i << 1 | 1 ; } while ( m -- ) { scanf ( a ) , scanf ( b ) ; addedge ( idx[a] , idx[b] ^ 1 ) ; addedge ( idx[b] , idx[a] ^ 1 ) ; } REP ( i , 0 , n << 1 ) if ( !dfn[i] ) tarjan ( i ) ; REP ( i , 0 , n ) if ( scc[i << 1] == scc[i << 1 | 1] ) return 0 ; return 1 ; } int main () { while ( ~scanf ( "%d%d" , &n , &m ) ) printf ( solve () ? "yes\n" : "no\n" ) ; return 0 ; }