HDU3062 Party (2-sat)

Party

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7029 Accepted Submission(s): 2299

Problem Description

有n对夫妻被邀请参加一个聚会，因为场地的问题，每对夫妻中只有1人可以列席。在2n 个人中，某些人之间有着很大的矛盾（当然夫妻之间是没有矛盾的），有矛盾的2个人是不会同时出现在聚会上的。有没有可能会有n 个人同时列席？

Input

n： 表示有n对夫妻被邀请 (n<= 1000)
m： 表示有m 对矛盾关系 ( m < (n - 1) * (n -1))

在接下来的m行中，每行会有4个数字，分别是 A1,A2,C1,C2
A1,A2分别表示是夫妻的编号
C1,C2 表示是妻子还是丈夫 ，0表示妻子 ，1是丈夫
夫妻编号从 0 到 n -1

Output

如果存在一种情况 则输出YES
否则输出 NO

Sample Input

2
1
0 1 1 1

Sample Output

YES

Source

2009 Multi-University Training Contest 16 - Host by NIT

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思路：2-sat裸题，直接根据关系建边然后判断

#include <iostream>  #include <cstdio>  #include <cstring>  #include <string>  #include <algorithm>  #include <cmath>  #include <map>  #include <set>  #include <stack>  #include <queue>  #include <vector>  #include <bitset>  using namespace std;#define LL long long  const int INF = 0x3f3f3f3f;#define MAXN 100100  #define MAXM 5000100  struct node{    int u, v, next;} edge[MAXM];int dfn[MAXN], low[MAXN], s[MAXN], Stack[MAXN], in[MAXN], block[MAXN];int cnt, tot, index, bnum, ctt;void init(){    memset(s, -1, sizeof s);    memset(dfn, 0, sizeof dfn);    memset(low, 0, sizeof low);    memset(Stack, 0, sizeof Stack);    memset(in, 0, sizeof in);    memset(block, 0, sizeof block);    cnt = tot = index = bnum = 0;}void add(int u, int v){    edge[cnt].u = u;    edge[cnt].v = v;    edge[cnt].next = s[u];    s[u] = cnt++;}void tarjan(int x){    dfn[x] = low[x] = ++tot;    Stack[++index] = x;    in[x] = 1;    for (int i = s[x]; ~i; i = edge[i].next)    {        int v = edge[i].v;        if (!dfn[v])        {            tarjan(v);            low[x] = min(low[x], low[v]);        }        else if (in[v])        {            low[x] = min(low[x], low[v]);        }    }    if (dfn[x] == low[x])    {        bnum++;        do        {            //printf("%d ",Stack[index]);              block[Stack[index]] = bnum;            in[Stack[index--]] = 0;        } while (Stack[index + 1] != x);        // printf("\n");      }}bool solve(){    for (int i = 0; i < ctt; i++)        if (!dfn[i])            tarjan(i);    for (int i = 0; i < ctt; i += 2)    {        if (block[i] == block[i ^ 1])            return 0;    }    return 1;}int main(){    int n,m,a,b,c,d;    while (~scanf("%d%d", &n, &m))    {        ctt = 2 * n;        init();        for (int i = 0; i < m; i++)        {            scanf("%d%d%d%d", &a, &b, &c, &d);            int x = 2 * a + c, y = 2 * b + d;            add(x, y ^ 1);            add(y, x ^ 1);        }        if (solve()) printf("YES\n");        else printf("NO\n");    }    return 0;}