la3211

Posted 2020-09-18 123456

2-sat+二分。。。

每次二分答案然后连边2-sat。。。边要开到n*n

样例水得跟没有一样。。。

#include<bits/stdc++.h>
using namespace std;
const int N = 4010;
struct edge {
    int nxt, to;
} e[N * N << 1];
int n, cnt = 1, top, Time, cot;
int dfn[N], low[N], vis[N], st[N], early[N], late[N], head[N], belong[N];
void link(int u, int v)
{
    e[++cnt].nxt = head[u];
    head[u] = cnt;
    e[cnt].to = v;
}
void tarjan(int u)
{
    st[++top] = u; dfn[u] = low[u] = ++Time; vis[u] = 1;
    for(int i = head[u]; i; i = e[i].nxt) 
    {
        if(!dfn[e[i].to])
        {
            tarjan(e[i].to);
            low[u] = min(low[u], low[e[i].to]);
        }
        else if(vis[e[i].to]) low[u] = min(low[u], dfn[e[i].to]);        
    }
    if(dfn[u] == low[u])
    {
        ++cot; int x = 0;
        while(x != u)
        {
            x = st[top--];
            belong[x] = cot;
            vis[x] = 0; 
        }
    }
}
bool judge(int t)
{
    memset(head, 0, sizeof(head));
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    memset(belong, 0, sizeof(belong));
    cnt = 1; top = Time = cot = 0;
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j) if(i != j)
        {
            int x = i << 1, y = j << 1;
            if(late[i] - late[j] >= 0 && late[i] - late[j] < t)
                link(x, y - 1), link(y, x - 1);
            if(late[i] - early[j] >= 0 && late[i] - early[j] < t)
                link(x, y), link(y - 1, x - 1);            
            if(early[i] - late[j] >= 0 && early[i] - late[j] < t)
                link(x - 1, y - 1), link(y, x);            
            if(early[i] - early[j] >= 0 && early[i] - early[j] < t)
                link(x - 1, y), link(y - 1, x);        
        }
    for(int i = 1; i <= 2 * n; ++i) if(!dfn[i]) tarjan(i);
    for(int i = 1; i <= n; ++i) if(belong[i * 2] == belong[i * 2 - 1]) return false;
    return true;
}
int main()
{
    while(scanf("%d", &n) != EOF)
    {
        int l = 0, r = 0, ans = 0;
        for(int i = 1; i <= n; ++i) scanf("%d%d", &early[i], &late[i]), r = max(r, late[i]);
        while(r - l > 1)
        {
            int mid = (l + r) >> 1;
            if(judge(mid)) ans = l = mid; else r = mid; 
        }
        printf("%d\n", ans);
    }
    return 0;
}

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