POJ 2749 二分+2-sat判定

最新推荐文章于 2021-02-26 21:51:37 发布

最新推荐文章于 2021-02-26 21:51:37 发布 · 96 阅读

本文探讨了一种通过构建图并应用二分判定方法来解决特定问题的算法策略。具体步骤包括引入必要的头文件，定义变量、数据结构如向量、栈和队列，并实现初始化、拓扑排序、构建图以及检查图是否满足二分条件的函数。通过输入实例数据，演示了如何使用这些函数来判断给定的图是否可以进行有效的二分划分。

这题还是挺好想建图的二分判定就行了

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <stack>
#include <queue>
#define MAXN 2005
#define MAXM 50005
#define INF 1000000000
using namespace std;
vector<int>g[MAXN];
int n, A, B;
int hatex[MAXN], hatey[MAXN], frix[MAXN], friy[MAXN];
int scc, index;
int low[MAXN], dfn[MAXN], instack[MAXN], fa[MAXN];
stack<int>st;
struct P
{
    int x, y;
}s1, s2, p[MAXN];
void init()
{
    for(int i = 0; i < MAXN; i++) g[i].clear();
    memset(dfn, 0, sizeof(dfn));
    while(!st.empty()) st.pop();
    memset(instack, 0, sizeof(instack));
    scc = index = 0;
}
void tarjan(int u)
{
    dfn[u] = low[u] = ++index;
    instack[u] = 1;
    st.push(u);
    int v;
    int size = g[u].size();
    for(int i = 0; i < size; i++)
    {
        v = g[u][i];
        if(!dfn[v])
        {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if(instack[v]) low[u] = min(low[u], dfn[v]);
    }
    if(dfn[u] == low[u])
    {
        scc++;
        do
        {
            v = st.top();
            st.pop();
            fa[v] = scc;
            instack[v] = 0;
        }while(v != u);
    }
}
int dist(P a, P b)
{
    return abs(a.x - b.x) + abs(a.y - b.y);
}
void build(int mid)
{
    for(int i = 1; i <= n; i++)
        for(int j = i + 1; j <= n; j++)
        {
            if(dist(p[i], s1) + dist(p[j], s1) > mid)
            {
                g[i].push_back(j + n);
                g[j].push_back(i + n);
            }
            if(dist(p[i], s2) + dist(p[j], s2) > mid)
            {
                g[i + n].push_back(j);
                g[j + n].push_back(i);
            }
            if(dist(s1, s2) + dist(p[i], s1) + dist(p[j], s2) > mid)
            {
                g[i].push_back(j);
                g[j + n].push_back(i + n);
            }
            if(dist(s1, s2) + dist(p[i], s2) + dist(p[j], s1) > mid)
            {
                g[i + n].push_back(j + n);
                g[j].push_back(i);
            }
        }
    for(int i = 1; i <= A; i++)
    {
        g[hatex[i]].push_back(hatey[i] + n);
        g[hatey[i]].push_back(hatex[i] + n);
        g[hatex[i] + n].push_back(hatey[i]);
        g[hatey[i] + n].push_back(hatex[i]);
    }
    for(int i = 1; i <= B; i++)
    {
        g[frix[i]].push_back(friy[i]);
        g[frix[i] + n].push_back(friy[i] + n);
        g[friy[i]].push_back(frix[i]);
        g[friy[i] + n].push_back(frix[i] + n);
    }
}
bool check()
{
    for(int i = 1; i <= 2 * n; i++)
        if(!dfn[i]) tarjan(i);
    for(int i = 1; i <= n; i++)
        if(fa[i] == fa[i + n]) return false;
    return true;
}
void solve()
{
    int low = 0, high = INF, ans = INF;
    while(low <= high)
    {
        int mid = (low + high) >> 1;
        init();
        build(mid);
        if(check()) {high = mid - 1; ans = min(ans, mid);}
        else low = mid + 1;
    }
    if(ans != INF)
    printf("%d\n", ans);
    else puts("-1");
}
int main()
{
    while(scanf("%d%d%d", &n, &A, &B) != EOF)
    {
        scanf("%d%d%d%d", &s1.x, &s1.y, &s2.x, &s2.y);
        for(int i = 1; i <= n; i++) scanf("%d%d", &p[i].x, &p[i].y);
        for(int i = 1; i <= A; i++) scanf("%d%d", &hatex[i], &hatey[i]);
        for(int i = 1; i <= B; i++) scanf("%d%d", &frix[i], &friy[i]);
        solve();
    }
    return 0;
}