POJ - 3905 Perfect Election(2-SAT)

来源：互联网发布：mysql #什么意思编辑：程序博客网时间：2024/04/30 17:26

题目大意：题意挺好理解的

解题思路：就按照题意来建边就可以了

#include <cstdio>#include <cstring>#include <vector>using namespace std;const int MAXNODE = 2010;vector<int> G[MAXNODE];bool mark[MAXNODE * 2];int Stack[MAXNODE * 2];int n, m, top;void AddClause(int x, int xval, int y, int yval) {    x = x * 2 + xval;    y = y * 2 + yval;    G[x].push_back(y);}void init() {    n++;    for (int i = 0; i < 2 * n; i++)        G[i].clear();    memset(mark, 0, sizeof(mark));    char s1[15], s2[15];    char sign1, sign2;    int x, valx, y, valy;    for (int i = 0; i < m; i++) {        scanf("%s%s", s1, s2);        sscanf(s1, "%c%d", &sign1, &x);        sscanf(s2, "%c%d", &sign2, &y);        valx = (sign1 == '+' ? 1 : 0);        valy = (sign2 == '+' ? 1 : 0);        AddClause(x, valx ^ 1, y, valy);        AddClause(y, valy ^ 1, x, valx);    }}bool dfs(int u) {    if (mark[u ^ 1]) return false;    if (mark[u]) return true;    mark[u] = true;    Stack[++top] = u;    for (int i = 0; i < G[u].size(); i++)        if (!dfs(G[u][i])) return false;    return true;}void solve() {    for (int i = 2; i < 2 * n; i += 2) {        if (!mark[i] && !mark[i^1]) {            top = 0;            if (!dfs(i)) {                while (top) mark[Stack[top--]] = false;                if (!dfs(i ^ 1)) {                    printf("0\n");                    return ;                }            }        }    }    printf("1\n");}int main() {    while (scanf("%d%d", &n, &m) != EOF) {        init();        solve();    }    return 0;}

0 0