题目:有一场婚礼,有n对夫妇参加,他们之间有些人之间有奸情(可能同性),在场的人中有一个公主,
她清楚其他人的人际关系,问能否安排座位使得两边都是n个人,且公主看不见有奸情的人同时在的对面。
分析:2-SAT。直接按照看的流程敲的程序。
1.建图,矛盾的点建立对应的边(与一直关系相反);
2.利用Tarjan算法计算强连通分量,缩点;
3.判断是否有解(是否有夫妇在同一集合);
4.如果成立,利用拓扑排序求出一组解,输出。
说明:第一个2SAT(⊙_⊙)。
#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> using namespace std; //link_list_begin typedef struct enode { int point; enode* next; }edge; edge*H1[204],*H2[204]; edge E1[40004],E2[40004]; int e_size1,e_size2; void link_init( void ) { e_size1 = 0; e_size2 = 0; memset( H1, 0, sizeof(H1) ); memset( H2, 0, sizeof(H2) ); memset( E1, 0, sizeof(E1) ); memset( E2, 0, sizeof(E2) ); } void link_add1( int a, int b ) { E1[e_size1].point = b; E1[e_size1].next = H1[a]; H1[a] = &E1[e_size1 ++]; } void link_add2( int a, int b ) { E2[e_size2].point = b; E2[e_size2].next = H2[a]; H2[a] = &E2[e_size2 ++]; } //link_list_end int s_id,times,top; int dfn[204],low[204],use[204],stk[204],oth[204],set[204]; void tarjan( int i, int j ) { dfn[i] = low[i] = ++ times; use[i] = 1; stk[++ top] = i; for ( edge* p = H1[i] ; p ; p = p->next ) { if ( !dfn[p->point] ) { tarjan(p->point, 0); low[i] = min(low[i], low[p->point]); }else if ( use[p->point] ) low[i] = min(low[i], dfn[p->point]); } if ( dfn[i] == low[i] ) { s_id ++; do { j = stk[top --]; use[j] = 0; set[j] = s_id; }while ( j != i ); } } void dfs( int i ) { for ( edge* p = H2[i] ; p ; p = p->next ) if ( !use[p->point] ) { dfs(p->point); use[i] = use[p->point]; use[oth[i]] = use[oth[p->point]]; } use[i] = 1; use[oth[i]] = 2; } void solve( int n ) { //强连通分量 s_id = times = top = 0; memset( dfn, 0, sizeof(dfn) ); memset( use, 0, sizeof(use) ); for ( int i = 0 ; i < 2*n ; ++ i ) if ( !dfn[i] ) tarjan(i, 0); //无解判断 for ( int i = 0 ; i < n ; ++ i ) if ( set[i*2] == set[i*2+1] ) { printf("bad luck\n"); return; }else { oth[set[i*2]] = set[i*2+1]; oth[set[i*2+1]] = set[i*2]; } //拓扑排序 times = 0; memset( use, 0, sizeof(use) ); for ( int i = 0 ; i < 2*n ; ++ i ) for ( edge* p = H1[i] ; p ; p = p->next ) if ( set[p->point] != set[i] ) link_add2( set[i], set[p->point] ); for ( int i = 1 ; i <= s_id ; ++ i ) if ( !use[i] ) dfs(i); for ( int i = 1 ; i < n ; ++ i ) { if ( i > 1 ) printf(" "); if ( use[set[i*2]] == use[set[0]] ) printf("%dw",i); else printf("%dh",i); } printf("\n"); } int main() { int n,m,x,y; char ch1,ch2; while ( ~scanf("%d%d",&n,&m) && n ) { link_init(); link_add1( 0, 1 ); for ( int i = 0 ; i < m ; ++ i ) { scanf("%d%c %d%c",&x,&ch1,&y,&ch2); x <<= 1; y <<= 1; if ( ch1 == 'w' ) x ^= 1; if ( ch2 == 'w' ) y ^= 1; link_add1( x^1, y ); link_add1( y^1, x ); } solve( n ); } return 0; }
UVa 11294 - Wedding
时间: 2024-10-12 09:33:35