LA 3211 Now or later (2-SAT + 二分)

来源：互联网发布：linux pid kill 编辑：程序博客网时间：2024/05/21 22:22

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1212

题意：n 架飞机，每架飞机的降落时间有两个选择，一个为早点降落，另一个为晚点降落，求两架飞机降落时间差的最小值的最大值（就是使得所有飞机降落的时间差的最小值尽量大）

思路：对时间进行二分，然后进行2-SAT的判断

#include <iostream>#include <queue>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int maxn = 2005;const int VN = maxn * 2;const int EN = maxn * maxn * 4;int n, m;int t[maxn][2];struct Graph{    int cnt, head[VN];    struct    {        int v, next;    } E[EN];    void init()    {        cnt = 0;        memset(head, -1, sizeof(head));    }    void add(int u, int v)    {        E[cnt].v = v;        E[cnt].next = head[u];        head[u] = cnt++;    }} g;struct Two_Sat{    bool check()    {        scc();        for(int i = 0; i < n; ++i)            if(col[i * 2] == col[i * 2 + 1])                return false;        return true;    }    void tarjan(const int u)    {        int v;        low[u] = dfn[u] = ++idx;        vis[u] = true;        sta[top++] = u;        for(int e = g.head[u]; e != -1; e = g.E[e].next)        {            v = g.E[e].v;            if(dfn[v] == -1)            {                tarjan(v);                low[u] = min(low[u], low[v]);            }            else                if(vis[v])                    low[u] = min(low[u], dfn[v]);        }        if(dfn[u] == low[u])        {            ++sn;            do            {                v = sta[--top];                vis[v] = false;                col[v] = sn;            }            while(u != v);        }    }    void scc()    {        idx = top = sn = 0;        memset(vis, 0, sizeof(vis));        memset(dfn, -1, sizeof(dfn));        for(int i = 0; i < 2 * n; ++i)            if(dfn[i] == -1)                tarjan(i);    }    int idx, top, sn;    int sta[VN], low[VN], dfn[VN], col[VN];    bool vis[VN];} sat;void build(int Min){    g.init();    for(int i = 0; i < n; ++i)    {        for(int j = i + 1; j < n; ++j)        {            int ta1 = t[i][0], ta2 = t[i][1];            int tb1 = t[j][0], tb2 = t[j][1];            if(abs(ta1 - tb1) < Min)            {                g.add(i * 2, j * 2 + 1);                g.add(j * 2, i * 2 + 1);            }            if(abs(ta1 - tb2) < Min)            {                g.add(i * 2, j * 2);                g.add(j * 2 + 1, i * 2 + 1);            }            if(abs(ta2 - tb1) < Min)            {                g.add(i * 2 + 1, j * 2 + 1);                g.add(j * 2, i * 2);            }            if(abs(ta2 - tb2) < Min)            {                g.add(i * 2 + 1, j * 2);                g.add(j * 2 + 1, i * 2);            }        }    }}int main(){    while(~scanf("%d", &n))    {        int l = 0, r = 0, mid, ans = -1;        for(int i = 0; i < n; ++i)        {            scanf("%d%d", &t[i][0], &t[i][1]);            r = max(t[i][0], max(t[i][1], r));        }        while(l < r)        {            mid = (l + r) >> 1;            build(mid);            if(sat.check())            {                ans = mid;                l = mid + 1;            }            else                r = mid;        }        printf("%d\n", ans);    }    return 0;}

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