gym 101485E 二分匹配

gym 101485E

题意：
给出 n 个 a,b 对，有三种运算符 + 、- 、* 。要你给出每对 a, b 指定运算符，使得最后 n个答案都不相同。
tags：
真是该退役了，写个二分匹配都写这么久 -_-
离散化，建个图跑匹配就好了。。也可以网络流

// gym 101485E
#include<bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
#define rep(i,a,b) for (int i=a; i<=b; ++i)
#define per(i,b,a) for (int i=b; i>=a; --i)
#define mes(a,b)  memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define MP make_pair
#define PB push_back
#define fi  first
#define se  second
typedef long long ll;
const int N = 2505, M = 1000005;
const ll  inf = 1e13;

int n;
ll  a[N], b[N];
vector< int > G[M];
map< pair<int , int > , pair<char, ll>  > mp;

int match[M];  bool used[M];
bool dfs(int u)
{
    used[u] = true;
    for(int to : G[u])
        if(match[to]==-1 || !used[match[to]] && dfs(match[to])) {
            match[u] = to,  match[to] = u;
            return true;
        }
    return false;
}
bool solve()
{
    memset(match, -1, sizeof(match));
    for(int i=1; i<=n; ++i)
        if(match[i]==-1) {
            memset(used, false, sizeof(used));
            if(!dfs(i)) return false;
        }
    return true;
}

ll  c[N];  int cnt=0;
int get_id(ll x) {
    return lower_bound(c+1, c+1+cnt, x) - c;
}
int main()
{
    scanf("%d", &n);
    rep(i,1,n) {
        scanf("%lld%lld", &a[i], &b[i]);
        c[++cnt] = a[i]+b[i]+inf;
        c[++cnt] = a[i]-b[i]+inf;
        c[++cnt] = a[i]*b[i]+inf;
        c[++cnt] = i;
    }
    sort(c+1, c+1+cnt);
    cnt = unique(c+1, c+1+cnt) - (c+1);
    rep(i,1,n) {
        int tmp = get_id(a[i]+b[i]+inf);   // +
        G[tmp].PB(i),  G[i].PB(tmp);
        mp[MP(i, tmp)] = MP('+', a[i]+b[i]);
        tmp = get_id(a[i]-b[i]+inf);        //-
        G[tmp].PB(i),  G[i].PB(tmp);
        mp[MP(i, tmp)] = MP('-', a[i]-b[i]);
        tmp = get_id(a[i]*b[i]+inf);      //*
        G[tmp].PB(i),  G[i].PB(tmp);
        mp[MP(i, tmp)] = MP('*', a[i]*b[i]);
    }
    if(!solve()) puts("impossible");
    else {
        rep(i,1,n) {
            int to = match[i];
            printf("%lld %c %lld = %lld
", a[i], mp[MP(i, to)].fi, b[i], mp[MP(i,to)].se);
        }
    }

    return 0;
}