UVa 11294 Wedding (TwoSat)

题意：有 n-1 对夫妻参加一个婚宴，所有人都坐在一个长长的餐桌左侧或者右侧，新郎和新娘面做面坐在桌子的两侧。由于新娘的头饰很复杂，她无法看到和她坐在同一侧餐桌的人，只能看到对面餐桌的人。任意一对夫妻不能坐在桌子的同侧，另外有m对人吵过架，而新娘不希望看到两个吵过架的人坐在他的对面，问如何安排这些座位。

析：很明显的TwoSat问题，假设mark[i<<1]标记了，就是和新娘同侧，否则就是和新娘对侧，这样的话，对于每对吵架的人，他们要么在不同的侧，要么在新娘同侧，注意一定要注意要提前把新娘标记了，而且题目有可能是新娘或者新郎吵过架，RE了好多次，还以为是数组开小了。。。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#include <numeric>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<LL, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e17;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 60 + 100;
const int maxm = 1e6 + 5;
const int mod = 10007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
}

struct TwoSat{
int n;
vector<int> G[maxn<<1];
bool mark[maxn<<1];
int S[maxn<<1], c;

void init(int n){
this-> n = n;
for(int i = 0; i < (n<<1); ++i)  G[i].cl;
ms(mark, 0);  mark[0] = 1;
}

void add_clause(int x, int xval, int y, int yval){
x = x << 1 | xval;
y = y << 1 | yval;
G[x^1].pb(y);
G[y^1].pb(x);
}

bool dfs(int x){
if(mark[x^1])  return false;
if(mark[x])  return true;
mark[x] = true;
S[c++] = x;
for(int i = 0; i < G[x].sz; ++i)
if(!dfs(G[x][i]))  return false;
return true;
}

bool solve(){
for(int i = 0; i < (n<<1); i += 2)
if(!mark[i] && !mark[i^1]){
c = 0;
if(!dfs(i)){
while(c > 0)  mark[S[--c]] = 0;
if(!dfs(i^1))  return false;
}
}
return true;
}
};
TwoSat twosat;

int main(){
while(scanf("%d %d", &n, &m) == 2 && n+m){
twosat.init(n);
int x, y;
char c1, c2;
while(m--){
scanf("%d%c %d%c", &x, &c1, &y, &c2);
twosat.add_clause(x, c1 == 'h', y, c2 == 'h');
}
if(!twosat.solve()){ puts("bad luck");  continue;  }
for(int i = 1; i < n; ++i)
printf("%d%c%c", i, "hw"[twosat.mark[i<<1]], " \n"[i+1==n]);
}
return 0;
}