南理工校赛I题

分析：

对2_SAT问题还是不熟，比赛的时候感觉像是2_SAT问题，也向这个方向想了，但是没想出来（好菜）。

2_SAT问题核心还是建图：比如这题，怎么建图？对于第$i$层，$x[i],y[i]$需要连接的边是$2*n-1-x[i]\to{y[i]},2*n-1-y[i]\to{x[i]}$，然后跑一下图看看是否$x[i]$和$2*n-1-x[i]$在同一个联通块。

然后二分答案，搞定！！!

代码：

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;
typedef vector <int>    VI;
typedef pair <int,int>  PII;
#define FOR(i,x,y)  for(int i = x;i < y;++ i)
#define IFOR(i,x,y) for(int i = x;i > y;-- i)
#define pb  push_back
#define mp  make_pair
#define fi  first
#define se  second

const int maxn = 2020;

VI  G[maxn];

int dfn[maxn],low[maxn],belong[maxn],mark[maxn],dfs_clock;
stack <int> s;

void dfs(int u){
    dfn[u] = low[u] = ++dfs_clock;
    s.push(u);
    mark[u] = 1;
    FOR(i,0,(int)G[u].size()){
        int v = G[u][i];
        if(!dfn[v]){
            dfs(v);
            low[u] = min(low[u],low[v]);
        }
        else if(mark[v])    low[u] = min(low[u],dfn[v]);
    }
    if(dfn[u] == low[u]){
        while(!s.empty()){
            int v = s.top();    s.pop();
            belong[v] = u;
            mark[v] = 0;
            if(v == u)  break;
        }
    }
}

int n;
PII point[maxn];

bool solve(){
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(mark,0,sizeof(mark));
    dfs_clock = 0;
    FOR(i,0,2*n)    if(!dfn[i]) dfs(i);
    FOR(i,0,2*n)    if(belong[i] == belong[2*n-1-i])    return false;
    return true;
}

void work(){
    int L = 0,R = n;
    while(L < R){
        int mid = (L+R+1)>>1;
        FOR(i,0,n<<1)   G[i].clear();
        FOR(i,0,mid)    G[2*n-1-point[i].fi].pb(point[i].se),
                        G[2*n-1-point[i].se].pb(point[i].fi);
        if(solve()) L = mid;
        else    R = mid-1;
    }
    printf("%d\n",L);
}

int main(){
    //freopen("test.in","r",stdin);
    int T;  scanf("%d",&T);
    while(T--){
        scanf("%d",&n);
        FOR(i,0,n)  scanf("%d%d",&point[i].fi,&point[i].se);
        work();
    }
    return 0;
}