参考题解

a,b存数组并且使用type记录当前位置选的是a里面的还是b里面的。这里0为a，1为b。 每次进行修改的时候 首先清除当前位置的最小累加，修改完成之后再加上即可。

#include <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <deque>
#include <unordered_map>
#include <map>
#include <cstring>
#include <cmath>
#include <unordered_set>
#include <set>
#include <utility>
#include <climits>
#include <iomanip>
#include <stack>
#include <bitset>

#define int long long
#define PII pair<int, int> 
#define TLLL tuple<int , int , int>
#define INF 0x3f3f3f3f3f3f3f3f
#define inf 0x3f
#define all(v) v.begin() + 1 , v.end()
#define ALL(v) v.begin() , v.end()
#define endl "\n"
using namespace std;

void solve()
{
    int n , q;
    cin >> n >> q;
    vector<int> a(n + 1) , b(n + 1) , type(n + 1);

    for (int i = 1 ; i <= n ; i ++ ) cin >> a[i];
    for (int i = 1 ; i <= n ; i ++ ) cin >> b[i];

    int mina = 0 , minb = 0;

    for (int i = 1 ; i <= n ; i ++ )
        if (a[i] <= b[i]) mina += a[i] , type[i] = 0;
        else minb += b[i] , type[i] = 1; 

    while (q -- )
    {
        char c;
        int x , y;
        cin >> c >> x >> y;

        if (type[x] == 0) mina -= a[x];
        else minb -= b[x];

        if (c == 'A') a[x] = y;
        else b[x] = y;

        if (a[x] <= b[x])
            type[x] = 0 , mina += a[x];
        else type[x] = 1 , minb += b[x];


        
        cout << mina + minb << endl;
    }

    return ;
}

signed main() 
{
    ios::sync_with_stdio(false);
    cin.tie(nullptr), cout.tie(nullptr);
    int T = 1;
    // cin >> T;
    while (T -- ) solve();
    return 0;
}