UVA11294-Wedding (2-Sat)

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N couples attend a wedding banquet. Everyone is sitting on the left or right side of a long dining table, and 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. There are more than m people, and the bride does not want to see them sitting on the same side. Ask if there is any allocation plan that meets the requirements of the bride.

Idea: 2-sat. Assume that each couple is a variable Xi. If Xi is true, the wife and the bride sit on the same side; if Xi is false, the husband and the bride sit on the same side. When Xi and XJ are husband ~ Xi v ~ XJ indicates that at most one of the two husbands is sitting on different sides of the bride. When Xi and XJ are both wives, they must satisfy Xi v XJ, it indicates that at most one of the two wives is sitting on different sides of the bride. When Xi and XJ are opposite sex ~ XI v XJ or xi v ~ One of XJ indicates that one of the two is sitting at most on different sides of the bride. To sum up, we need to satisfy our husband ~ XI, wife Xi. At last, pay attention to initializing 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;}

UVA11294-Wedding (2-Sat)