UVA11294-Wedding(2-SAT)

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The question ： Yes n The couple attended a wedding banquet .

Everyone sits on the left or right of a long table . All couples can only sit face to face , Including the bride and groom . The bride can only see people sitting on different sides of her . Yes m It's more than a fight for people , The bride doesn't want to see them sitting on the same side .

Ask if there is a distribution plan to meet the bride's requirements .

Ideas ：2-SAT problem . If each couple is a variable xi. If xi by true when , The wife and the bride sit on the same side ;xi by false when , The husband and the bride sit on the same side . When xi and xj When you are both husband , It needs to meet ~xi V ~xj, It means that at most one of the two husbands is sitting on a different side from the bride ; When xi and xj When both are wives . It needs to meet xi V xj. It means that at most only one of the two wives is sitting on a different side from the bride . When xi and xj When opposite sex , It needs to meet ~xi V xj perhaps xi V ~xj One of them . It means that at most one of the two is sitting on a different side from the bride .

in summary . Is to satisfy the husband ~xi, Wife xi. Finally, pay attention to initialization mark[1] = 1.

Code ：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>

using namespace std;

const int MAXN = 1005;

struct TwoSAT{
    int n;
    vector<int> g[MAXN * 2];
    bool mark[MAXN * 2];
    int s[MAXN * 2], c;

    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].size(); i++)
            if (!dfs(g[x][i])) return false;
        return true;
    }

    void init(int n) {
        this->n = n;
        for (int i = 0; i < n * 2; i++) 
            g[i].clear();
        memset(mark, 0, sizeof(mark));
        mark[1] = 1;
    }

    void add_clause(int x, int xval, int y, int yval) {
        x = x * 2 + xval;
        y = y * 2 + yval;
        g[x^1].push_back(y);
        g[y^1].push_back(x);
    }

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

TwoSAT solver;

int n, m;

int main() {
    while (scanf("%d%d", &n, &m)) {
        if (n == 0 && m == 0)
            break;
        solver.init(n);         
        char a, b;
        int xval, yval, u, v;
        while (m--) {
            scanf("%d%c%d%c", &u, &a, &v, &b); 
            xval = (a == 'h') ? 0 : 1;            yval = (b == 'h') ? 0 : 1;            solver.add_clause(u, xval, v, yval);        }        if (!solver.solve())            printf("bad luck\n");        else {            for (int i = 1; i < n; i++) {                printf("%d%c", i, solver.mark[2*i] ? 'h' : 'w');                if (i == n - 1) printf("\n");                else printf(" ");            }         }    }    return 0;}

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