POJ-3648（2-SAT）

题目：http://poj.org/problem?id=3648

第四道2-SAT，本来以为很简单的，没想到WA了一个晚上~~~~(>_<)~~~~ 。刚开始没考虑到bride和groom的通奸情况，然后想到到了又考虑错了以为bride不能看到那个情夫，看了discuss才知道原来bride看到情夫也没关系，毕竟bride看不到自己，真是给跪了。。。

#include <cstdio>#include <cctype>#include <cstring>#include <vector>#include <algorithm>using namespace std;#define MAX_N 1005int N, M;bool predefined[MAX_N * 2];//those who must sit at bride's sidevector<int> out[MAX_N * 4];bool mark[MAX_N * 4];//mark[2i] = 1, i sit at bride side; mark[2i+1] = 1, i sit at groom sideint stack[MAX_N * 4], cnt;inline void addEdge(int i, int j){//i can not sit in the same side with jint x =  i << 1, y = j << 1;out[x].push_back(y+1);out[y+1].push_back(x);out[x+1].push_back(y);out[y].push_back(x+1);}inline void arrange(int i, int j){if(i > j) swap(i, j);int x = i << 1, y = j << 1;if(i == 0){//j must seat at bride's side so that bride can not see herpredefined[j] = true;}else if(i == 1){//it does not matter as the bride can no see herself}else{//i and j can not both sit at groom's sideout[x+1].push_back(y);out[y+1].push_back(x);}}inline int IDtoInt(const char* s){//husbands' ids are even, wives' ids are oddint id = 0;for(; isdigit(*s); ++s) id = id * 10 + *s - '0';if(*s == 'h') return id << 1;return id * 2 + 1;}inline char* IntToID(char* s, int n){if(n & 1) sprintf(s, "%dw", n >> 1);else sprintf(s, "%dh", n >> 1);return s;}void init(){memset(predefined, 0, N<<1);memset(mark, 0, N<<2);for(int i = N*4-1; i > -1; --i) out[i].clear();}void build(){int i, x;char l[8], r[8];for(i = 0; i < N; ++i){x = i << 1;addEdge(x, x + 1);}for(i = 0; i < M; ++i){//those can not sit at the same tablescanf("%s%s", l, r);arrange(IDtoInt(l), IDtoInt(r));}}bool dfs(int x){if(mark[x ^ 1]) return false;if(mark[x]) return true;mark[x] = true;stack[cnt++] = x;const vector<int>& v = out[x];for(int i = v.size() - 1; i > -1; --i){if(!dfs(v[i])) return false;}return true;}bool test(){int i, n = N << 1;if(!dfs(0 << 1 | 1)) return false;//groom must sit at groom's sideif(!dfs(1 << 1)) return false;//bride must sit at bride's side//those predefined must sit at bride's sidefor(i = 2; i < n; ++i){if(predefined[i] && !dfs(i << 1)) return false;}//normal 2-SATfor(i = 2; i < n; ++i){cnt = 0;if(!dfs(i << 1)){while(cnt) mark[stack[--cnt]] = false;if(!dfs(i << 1 | 1)) return false;}}return true;}void print(){int i, j, n = N << 1, sum = 0;char s[8];bool first = true;for(i = 2; i < n && sum < N; i += 2){//check if the husband or the wife should sit at the bride's sidej = i;if(!mark[i << 1]) j = i ^ 1;++sum;if(first) first = false;else putchar(' ');printf("%s", IntToID(s, j));}puts("");}int main(){while(scanf("%d%d", &N, &M), N){init();build();if(test()) print();else puts("bad luck");}return 0;}