POJ 3678 Katu Puzzle

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大意不再赘述。

思路：有几个表达式我们可以假设c的值为0或者1，然后去推相互之间有矛盾的点，连边时，我们将点拆为2个，一个表示0，一个表示1，Tarjan时就需要循环到2*n啦，有些表达式重复了，比如异或所有的重复了，所以不需要再增加边。

a AND b == 1 (a,b同时为1) a->b, b->a, ~a-->a, ~b-->b;
a AND b == 0 (a,b不同时为1) a-->~b, b-->~a;
a OR b == 1 (a,b不同时为0) ~a-->b, ~b-->a;
a OR b == 0 (a,b同时为0) ~a->~b, ~b->~a, a-->~a, b-->~b;
a XOR b == 1 (a,b不相同为1) a->~b, b->~a, ~a->b, ~b->a;
a XOR b == 0 (a,b相同为0) a->b, b->a, ~a->~b, ~b->~a;

#include <iostream>#include <cstdio>#include <algorithm>#include <cstring>#include <string>#include <cstdlib>using namespace std;const int MAXN = 2020;const int MAXM = 1000100;struct Edge{    int u, v, next;}edge[MAXM];int first[MAXN], stack[MAXN], ins[MAXN], low[MAXN], dfn[MAXN];int belong[MAXM];int ind[MAXN], outd[MAXN];int cnt;int n, m;int scnt, top, tot;void init(){    cnt = 0;    scnt = top = tot = 0;    memset(first, -1, sizeof(first));    memset(ins, 0, sizeof(ins));    memset(dfn, 0, sizeof(dfn));}void read_graph(int u, int v){    edge[cnt].v = v;    edge[cnt].next = first[u], first[u] = cnt++;}void dfs(int u){    int v;    low[u] = dfn[u] = ++tot;    stack[top++] = u;    ins[u] = 1;    for(int e = first[u]; e != -1; e = edge[e].next)    {        v = edge[e].v;        if(!dfn[v])        {            dfs(v);            low[u] = min(low[u], low[v]);        }        else if(ins[v])        {            low[u] = min(low[u], dfn[v]);        }    }    if(low[u] == dfn[u])    {        scnt++;        do        {            v = stack[--top];            belong[v] = scnt;            ins[v] = 0;        }while(u != v);    }}void Tarjan(){    for(int v = 0; v < 2*n; v++) if(!dfn[v])        dfs(v);}void read_case(){init();while(m--){int x, y, c;char str[5];scanf("%d%d%d%s", &x, &y, &c, str);if(str[0] == 'A') //拆点为2*x, 2*x+1，前者表示~a,后者表示a {if(c){read_graph(2*x, 2*x+1);read_graph(2*y, 2*y+1);}else{read_graph(2*x+1, 2*y);read_graph(2*y+1, 2*x);}}else if(str[0] == 'O'){if(c){read_graph(2*x, 2*y+1);read_graph(2*y, 2*x+1);}else{read_graph(2*x+1, 2*x);read_graph(2*y+1, 2*y);}}}}void solve(){    read_case();    Tarjan();    for(int i = 0; i < 2*n; i += 2)    {        if(belong[i] == belong[i+1])        {            printf("NO\n");            return ;        }    }    printf("YES\n");    }int main(){    while(~scanf("%d%d", &n, &m))    {        solve();    }    return 0;}