hdu 3062 Party 最裸2-sat题目

http://acm.hdu.edu.cn/showproblem.php?pid=3062

题意：中文...

思路：

来自伍昱的《由对称性解2-SAT问题》

http://wenku.baidu.com/view/afd6c436a32d7375a41780f2.html

把确定不矛盾的双方建立边，然后tarjan缩点。判断每一对夫妻是否属于不同的环，如果存在同一环里，则无解，否则有解

//#pragma comment(linker,"/STACK:327680000,327680000")

#include <iostream>

#include <cstdio>

#include <cmath>

#include <vector>

#include <cstring>

#include <algorithm>

#include <string>

#include <set>

#include <functional>

#include <numeric>

#include <sstream>

#include <stack>

#include <map>

#include <queue>



#define CL(arr, val)    memset(arr, val, sizeof(arr))



#define inf 0x7f7f7f7f

#define lc l,m,rt<<1

#define rc m + 1,r,rt<<1|1

#define pi acos(-1.0)

#define ll long long

#define L(x)    (x) << 1

#define R(x)    (x) << 1 | 1

#define MID(l, r)   (l + r) >> 1

#define Min(x, y)   (x) < (y) ? (x) : (y)

#define Max(x, y)   (x) < (y) ? (y) : (x)

#define E(x)        (1 << (x))

#define iabs(x)     (x) < 0 ? -(x) : (x)int

#define OUT(x)  printf("%I64d\n", x)

#define lowbit(x)   (x)&(-x)

#define Read()  freopen("din.txt", "r", stdin)

#define Write() freopen("dout.txt", "w", stdout);



#define N 2005



using namespace std;





int dfn[N],low[N];

int belong[N],stk[N];

bool ins[N];

int cnt,idx,top;

vector<int>iVc[N];



int n,m;



void tarjan(int u)

{

    int j;

    dfn[u] = low[u] = ++idx;

    stk[++top] = u;

    ins[u] = true;

    int sz = iVc[u].size();

    for (j = 0; j < sz; ++j)

    {

        int v = iVc[u][j];

        if (dfn[v] == -1)

        {

            tarjan(v);

            low[u] = min(low[u],low[v]);

        }

        else if (ins[v])

        {

            low[u] = min(low[u],dfn[v]);

        }

    }

    if (dfn[u] == low[u])

    {

        cnt++;

        do

        {

            j = stk[top--];

            ins[j] = false;

            belong[j] = cnt;

        }while (j != u);

    }

}

void solve()

{

    int i;

    //tarjan缩点

    for (i = 0; i < 2*n; ++i)

    {

        if (dfn[i] == -1)

        {

            tarjan(i);

        }

    }

    //判断夫妻是否属于同一环

    bool flag = false;

    for (i = 0; i < n; ++i)

    {

        if (belong[2*i] == belong[2*i + 1])

        {

            flag = true;

            break;

        }

    }

    if (!flag)

    printf("YES\n");

    else

    printf("NO\n");

}

void init()

{

    int i;

    for (i = 0; i <= 2*n; ++i)

    {

        iVc[i].clear();

        dfn[i] = low[i] = -1;

        ins[i] = false;

    }

    top = idx = cnt = 0;

}

int main()

{

    int i;

    int a1,a2,b1,b2;

    while (~scanf("%d%d",&n,&m))

    {

        init();

        for (i = 0; i < m; ++i)

        {

            //不矛盾的建边

            scanf("%d%d%d%d",&a1,&a2,&b1,&b2);

            iVc[a1*2 + b1].push_back(a2*2 + 1 - b2);

            iVc[a2*2 + b2].push_back(a1*2 + 1 - b1);

        }

        solve();

    }

    return 0;

}