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Businessmen Problems time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output

Two famous competing companies ChemForces and TopChemist decided to show their sets of recently discovered chemical elements on an exhibition. However they know that no element should be present in the sets of both companies.

In order to avoid this representatives of both companies decided to make an agreement on the sets the companies should present. The sets should be chosen in the way that maximizes the total income of the companies.

All elements are enumerated with integers. The ChemForces company has discovered $\mathrm{nn\; distinct\; chemical\; elements\; with\; indices}$

a1,a2,…,ana1,a2,…,an, and will get an income of ${\mathrm{xixi\; Berland\; rubles\; if\; the\; ii-th\; element\; from\; this\; list\; is\; in\; the\; set\; of\; this\; company.}}^{}$

The TopChemist company discovered $\mathrm{mm\; distinct\; chemical\; elements\; with\; indices\; b1,b2,\dots ,bmb1,b2,\dots ,bm,\; and\; it\; will\; get\; an\; income\; of\; yjyj\; Berland\; rubles\; for\; including\; the}$

jj-th element from this list to its set.

In other words, the first company can present any subset of elements from $\{a1,a2,\dots ,an\}\{a1,a2,\dots ,an\}\; (possibly\; empty\; subset),\; the\; second\; company\; can\; present\; any\; subset\; of\; elements\; from\; \{b1,b2,\dots ,bm\}\{b1,b2,\dots ,bm\}\; (possibly\; empty\; subset).\; There\; shouldn‘t\; be\; equal\; elements\; in\; the\; subsets.$

Help the representatives select the sets in such a way that no element is presented in both sets and the total income is the maximum possible.

Input

The first line contains a single integer $\mathrm{nn\; (1\le n\le 1051\le n\le 105)\; —\; the\; number\; of\; elements\; discovered\; by\; ChemForces.}$

The $\mathrm{ii-th\; of\; the\; next\; nn\; lines\; contains\; two\; integers\; aiai\; and\; xixi\; (1\le ai\le 1091\le ai\le 109,\; 1\le xi\le 1091\le xi\le 109)\; —\; the\; index\; of\; the\; ii-th\; element\; and\; the\; income\; of\; its\; usage\; on\; the\; exhibition.\; It\; is\; guaranteed\; that\; all\; aiai\; are\; distinct.}$

The next line contains a single integer $\mathrm{mm\; (1\le m\le 1051\le m\le 105)\; —\; the\; number\; of\; chemicals\; invented\; by\; TopChemist.}$

The $\mathrm{jj-th\; of\; the\; next\; mm\; lines\; contains\; two\; integers\; bjbj\; and\; yjyj,\; (1\le bj\le 1091\le bj\le 109,\; 1\le yj\le 1091\le yj\le 109)\; —\; the\; index\; of\; the\; jj-th\; element\; and\; the\; income\; of\; its\; usage\; on\; the\; exhibition.\; It\; is\; guaranteed\; that\; all\; bjbj\; are\; distinct.}$

Output

Print the maximum total income you can obtain by choosing the sets for both companies in such a way that no element is presented in both sets.

Examples input Copy

3
1 2
7 2
3 10
4
1 4
2 4
3 4
4 4

output Copy

24

input Copy

1
1000000000 239
3
14 15
92 65
35 89

output Copy

408

Note

In the first example ChemForces can choose the set ($\mathrm{3,73,7),\; while\; TopChemist\; can\; choose\; (1,2,41,2,4).\; This\; way\; the\; total\; income\; is\; (10+2)+(4+4+4)=24(10+2)+(4+4+4)=24.}$

In the second example ChemForces can choose the only element ${\mathrm{109109,\; while\; TopChemist\; can\; choose\; (14,92,3514,92,35).\; This\; way\; the\; total\; income\; is\; (239)+(15+65+89)=408(239)+(15+65+89)=408.}}^{}$

題意： 給你n個A公司的新元素及價值，再給你m個B公司的新元素及價值，兩個公司的元素相同的取較大值，問兩家公司元素價值總和

直接用一個map保存下相同元素的較大值，然後遍歷相加就可以了

#include <map>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <cstdio>
#include <vector>
#include <string>
#include <cstring>
#include <iostream>
#include <algorithm>
#define debug(a) cout << #a << " " << a << endl
using namespace std;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
typedef long long ll;
int main(){
    std::ios::sync_with_stdio(false);
    ll n, m;
    while( cin >> n ) {
        map<ll,ll> mm;
        ll x, y;
        for( ll i = 0; i < n; i ++ ) {
            cin >> x >> y;
            if( mm[x] < y ) {
                mm[x] = y;
            }
        }
        cin >> m;
        for( ll i = 0; i < m; i ++ ) {
            cin >> x >> y;
            if( mm[x] < y ) {
                mm[x] = y;
            }
        }
        ll sum = 0;
        for( map<ll,ll>:: iterator it = mm.begin(); it != mm.end(); it ++ ) {
            sum += (*it).second;
        }
        cout << sum << endl;
    }
    return 0;
}

CF981B Businessmen Problems map 模擬 二十二