HDU3715 2-SAT

来源：互联网发布：sql中with as的用法编辑：程序博客网时间：2024/05/01 21:04
#include <iostream>#include <cstdio>#include <cstring>#include <vector>#include <stack>using namespace std;const int M = 20005;int dfn[M];int low[M];int ins[M];int sccf[M];int a[M];int b[M];int c[M];vector<int>que[M];stack<int>s;int index, scc;int n, m;int ans;void init() {    index = 1;    scc = 0;    memset(dfn, 0, sizeof(dfn));    memset(low, 0, sizeof(low));    memset(ins, 0, sizeof(ins));    memset(sccf, 0, sizeof(sccf));    for(int i = 0; i <= n*2; i++)        que[i].clear();    while(!s.empty())        s.pop();}/*假设a = 0 则 a‘ = 1， b = 0 则 b’ = 1；将a b 定义一种值， 则相应的值都将确定，如果这种情况可以满足题目要求， 这此题与解，因为题目中的数值都是相对的。如果 C = 0 , 则 a 和b 矛盾， 则a -> b' b ->a'如果 c = 1 则  a 和b' b 和a'矛盾 则 a -> b b' ->a' b ->a a' ->b'如果 c = 2 则  a‘ 和 b'矛盾， 则 a'->b b'->a*/void build_map(int mid) {    for(int i = 1; i <= mid; i++) {        if(c[i] == 0) {            que[a[i]].push_back(b[i] + n);            que[b[i]].push_back(a[i] + n);        }else if(c[i] == 1) {            que[a[i]].push_back(b[i]);            que[b[i] + n].push_back(a[i] + n);            que[b[i]].push_back(a[i]);            que[a[i] + n].push_back(b[i] + n);        }else {            que[a[i] + n].push_back(b[i]);            que[b[i]+n].push_back(a[i]);        }    }}void Tarjan(int u) {    int v;    dfn[u] = low[u] = index++;    ins[u] = 1;    s.push(u);    for(int i = 0; i <(int)que[u].size(); i++) {        v = que[u][i];        if(!dfn[v]) {            Tarjan(v);            low[u] = min(low[u], low[v]);        }else if(ins[v]) {            low[u] = min(low[u], low[v]);        }    }    if(dfn[u] == low[u]) {        scc++;        do {            v = s.top();            s.pop();            ins[v] = 0;            sccf[v] = scc;        }while(u != v);    }}bool judge(int mid) {     init();   build_map(mid);    for(int i = 0; i <2*n; i++) {        if(!dfn[i])            Tarjan(i);    }    for(int i = 0; i < n; i++)        if(sccf[i] == sccf[i + n]){           return false;        }    return true;}void work() {    int l = 0, r = m;    ans = 0;    while(l <= r) {        int mid = (l + r)/2;        if(judge(mid)) {          ans = mid;            l = mid+1;        }else {            r = mid -1;        }    }    printf("%d\n", ans);}int main(){    int t;    scanf("%d", &t);    while(t--){        scanf("%d%d", &n, &m);        for(int i = 1; i <= m; i++)           scanf("%d%d%d", &a[i], &b[i], &c[i]);       // printf("ddd\n");        work();    }    return 0;}
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