Uva11294-wedding (2-sat)

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Test Instructions: There are n couples participating in a wedding reception.

All of them sit on the left or right side of a long table. All couples are only able to sit in person, including the bride and groom. The bride can only see the people sitting on her different sides. There is M-to-man over the shelf, and the bride does not 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 is true, the wife and the bride sit on the same side; When Xi is false, the husband and the bride sit on the same side. When Xi and XJ are husbands, it is necessary to meet ~xi V ~XJ, which means that two husbands have at most only one sitting on a different side of the bride, while Xi and XJ are wives. The XI V XJ is to be met. Indicates that two wives have at most only one sitting on a different side from the bride. When Xi and XJ are heterosexual, one of the ~xi v XJ or XI V ~XJ is required. Indicates that two of the most are seated on a different side than the bride.

Sum up. is to satisfy the husband ~xi, wife XI. Finally, be aware of 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;}

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Uva11294-wedding (2-sat)