CF981B Businessmen Problems map 模拟 二十二

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 $\mathrm{nn distinct\; chemical\; elements\; with\; indices a1,a2,\dots ,ana1,a2,\dots ,an,\; and\; will\; get\; an\; income\; of xixi Berland\; rubles\; if\; the ii-th\; element\; from\; this\; list\; is\; in\; the\; set\; of\; this\; company.}$

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 $\{a1,a2,\dots ,an\}\{a1,a2,\dots ,an\} (possibly\; empty\; subset),\; the\; second\; company\; can\; present\; any\; subset\; of\; elements\; from \{b1,b2,\dots ,bm\}\{b1,b2,\dots ,bm\} (possibly\; empty\; subset).\; There\; shouldn\text{'}t\; be\; equal\; elements\; in\; the\; subsets.$

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

 

 题意： 给你n个A公司的新元素及价值，再给你m个B公司的新元素及价值，两个公司的元素相同的取较大值，问两家公司元素价值总和

直接用一个map保存下相同元素的较大值，然后遍历相加就可以了

#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;
}

 

转载于:https://www.cnblogs.com/l609929321/p/9250018.html