模型：o4-mini-high
思考持续时间：4分25秒
图片证据：

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AC代码：

#include <bits/stdc++.h>
using namespace std;
using pii = pair<int,int>;
using vi = vector<int>;
using ll = long long;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int T;
    cin >> T;
    while(T--){
        int n;
        cin >> n;
        string s1, s2, t1, t2;
        cin >> s1 >> s2 >> t1 >> t2;
        s1 = " " + s1;
        s2 = " " + s2;
        t1 = " " + t1;
        t2 = " " + t2;

        // prefix sums of ones for s1 and s2
        vector<int> p1(n+1, 0), p2(n+1, 0);
        for(int i = 1; i <= n; i++){
            p1[i] = p1[i-1] + (s1[i]=='1');
            p2[i] = p2[i-1] + (s2[i]=='1');
        }

        // events: at segment start i, add (end, count0, count1)
        vector<vector<array<int,3>>> ev1(n+2), ev2(n+2);

        // build s1 segments
        for(int i = 1; i <= n; ){
            if(t1[i]=='1'){
                int L = i;
                int j = i;
                while(j <= n && t1[j]=='1') j++;
                int R = j-1;
                int ones = p1[R] - p1[L-1];
                int len = R - L + 1;
                int zeros = len - ones;
                ev1[L].push_back({R, zeros, ones});
                i = R + 1;
            } else i++;
        }
        // build s2 segments
        for(int i = 1; i <= n; ){
            if(t2[i]=='1'){
                int L = i;
                int j = i;
                while(j <= n && t2[j]=='1') j++;
                int R = j-1;
                int ones = p2[R] - p2[L-1];
                int len = R - L + 1;
                int zeros = len - ones;
                ev2[L].push_back({R, zeros, ones});
                i = R + 1;
            } else i++;
        }

        // min-heaps of deadlines for supplies
        priority_queue<int, vector<int>, greater<int>> h1[2], h2[2];
        ll ans = 0;

        for(int i = 1; i <= n; i++){
            // 1) remove expired supplies
            for(int b = 0; b < 2; b++){
                while(!h1[b].empty() && h1[b].top() < i) h1[b].pop();
                while(!h2[b].empty() && h2[b].top() < i) h2[b].pop();
            }

            // 2) add new segment supplies
            for(auto &seg : ev1[i]){
                int R = seg[0], c0 = seg[1], c1 = seg[2];
                for(int k = 0; k < c0; k++) h1[0].push(R);
                for(int k = 0; k < c1; k++) h1[1].push(R);
            }
            for(auto &seg : ev2[i]){
                int R = seg[0], c0 = seg[1], c1 = seg[2];
                for(int k = 0; k < c0; k++) h2[0].push(R);
                for(int k = 0; k < c1; k++) h2[1].push(R);
            }

            // 3) process this position
            bool can1 = (t1[i]=='1');
            bool can2 = (t2[i]=='1');
            if(!can1 && !can2){
                // both fixed
                if(s1[i] == s2[i]) ans++;
            }
            else if(can1 && !can2){
                // s1 supply -> s2 fixed
                int bit2 = s2[i] - '0';
                if(!h1[bit2].empty()){
                    h1[bit2].pop();
                    ans++;
                } else if(!h1[1-bit2].empty()){
                    h1[1-bit2].pop();
                }
            }
            else if(!can1 && can2){
                // s2 supply -> s1 fixed
                int bit1 = s1[i] - '0';
                if(!h2[bit1].empty()){
                    h2[bit1].pop();
                    ans++;
                } else if(!h2[1-bit1].empty()){
                    h2[1-bit1].pop();
                }
            }
            else {
                // both supply: type3
                // try match same bits first
                if(!h1[0].empty() && !h2[0].empty()){
                    h1[0].pop();
                    h2[0].pop();
                    ans++;
                }
                else if(!h1[1].empty() && !h2[1].empty()){
                    h1[1].pop();
                    h2[1].pop();
                    ans++;
                }
                else {
                    // no direct match, consume one from each if available
                    if(!h1[0].empty() || !h1[1].empty()){
                        // pick earliest deadline to consume
                        int d0 = h1[0].empty() ? INT_MAX : h1[0].top();
                        int d1 = h1[1].empty() ? INT_MAX : h1[1].top();
                        if(d0 <= d1) h1[0].pop();
                        else h1[1].pop();
                    }
                    if(!h2[0].empty() || !h2[1].empty()){
                        int d0 = h2[0].empty() ? INT_MAX : h2[0].top();
                        int d1 = h2[1].empty() ? INT_MAX : h2[1].top();
                        if(d0 <= d1) h2[0].pop();
                        else h2[1].pop();
                    }
                }
            }
        }

        // finished this test
        cout << ans << "\n";
    }

    return 0;
}