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UVA11294-Wedding(2-SAT)
2022-07-05 23:03:00 【全栈程序员站长】
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题意:有n对夫妻參加一个婚宴。
全部人都坐在一个长长的餐桌的左边或者右边。全部夫妻都仅仅能面对面坐,包含新娘和新郎。新娘仅仅能看到坐在她不同側的人。有m对人超过架,新娘不希望看到他们坐在同一側。
问有没有分配方案满足新娘的要求。
思路:2-SAT问题。如果每对夫妇为一个变量xi。如果xi为true时,妻子与新娘坐同一側;xi为false时,丈夫与新娘坐同一側。当xi和xj同为丈夫时,则需满足~xi V ~xj,表示两个丈夫最多仅仅有一个坐在与新娘不同側;当xi和xj同为妻子时。则需满足xi V xj。表示两个妻子最多仅仅有一个坐在与新娘不同側。当xi和xj为异性时,则需满足~xi V xj或者xi V ~xj当中一个。表示两个最多就一个坐在与新娘不同側。
综上所述。就是要满足丈夫~xi,妻子xi。最后要注意初始化mark[1] = 1。
代码:
#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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