QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#874996#9345. Artful PaintingsSkyWaveWA 99ms3840kbC++143.1kb2025-01-28 23:27:472025-01-28 23:27:48

Judging History

你现在查看的是最新测评结果

  • [2025-01-28 23:27:48]
  • 评测
  • 测评结果:WA
  • 用时:99ms
  • 内存:3840kb
  • [2025-01-28 23:27:47]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;

void solve() {
    constexpr int N = 3000, M = 3000;
    
    static int edge[M + M + 1 + N + N + 1 + 1][3], tail[N + 1];
    
    int n, m1, m2;
    cin >> n >> m1 >> m2;
    
    memset(tail, 0, sizeof(int) * (n + 1));
    
    int m = 0;
    
    auto addEdge = [&](int u, int v, int w) {
        edge[++m][0] = v;
        edge[m][1] = w;
        edge[m][2] = tail[u];
        tail[u] = m;
    };
    
    for (int i = 1; i <= m1; ++i) {
        int l, r, k;
        cin >> l >> r >> k;
        addEdge(r, l - 1, -k);  // S[r] - S[l-1] >= k → S[l-1] <= S[r] - k
    }
    
    for (int i = 1; i <= m2; ++i) {
        int l, r, k;
        cin >> l >> r >> k;
        addEdge(l - 1, r, -k);  // 初始权值为 -k,后续调整时加上 mid
    }
    
    for (int i = 1; i <= n; ++i) {
        addEdge(i, i - 1, 0);       // S[i] >= S[i-1]
        addEdge(i - 1, i, 1);       // S[i] <= S[i-1] + 1
    }
    
    auto check = [&](int mid) {
        // 调整类型2约束的权值:边权变为 -k + mid
        for (int i = m1 + 1; i <= m1 + m2; ++i) {
            edge[i][1] += mid;
        }
        
        int old_m = m;
        // 添加两条边强制 S[n] = mid
        addEdge(0, n, mid);     // S[n] <= mid
        addEdge(n, 0, -mid);    // S[n] >= mid
        
        queue<int> que;
        que.push(0);
        static bitset<N + 1> inq;
        inq.reset();
        inq[0] = true;
        static int dis[N + 1];
        memset(dis + 1, 0x3f, sizeof(int) * n);
        dis[0] = 0;
        static int cnt[N + 1];
        memset(cnt, 0, sizeof(int) * (n + 1));
        
        bool res = true;
        while (!que.empty() && res) {
            int u = que.front(); que.pop();
            inq[u] = false;
            for (int i = tail[u]; i; i = edge[i][2]) {
                int v = edge[i][0], w = edge[i][1];
                if (dis[u] + w < dis[v]) {
                    dis[v] = dis[u] + w;
                    if (!inq[v]) {
                        if (++cnt[v] > n) {  // 包含n+1个节点,有负环
                            res = false;
                            break;
                        }
                        que.push(v);
                        inq[v] = true;
                    }
                }
            }
        }
        
        // 恢复边的状态
        m = old_m;
        for (int i = m1 + 1; i <= m1 + m2; ++i) {
            edge[i][1] -= mid;
        }
        // 恢复 tail 数组,因为添加的临时边被撤销
        tail[0] = edge[m][2]; // 恢复原来的最后一条边
        tail[n] = edge[m-1][2]; // 假设 n 的边在添加时没有其他操作
        
        return res;
    };
    
    int l = 0, r = n;
    while (l <= r) {
        int mid = (l + r) >> 1;
        if (check(mid)) {
            r = mid - 1;
        } else {
            l = mid + 1;
        }
    }
    
    cout << l << '\n';
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int t;
    cin >> t;
    while (t--) {
        solve();
    }
    return 0;
}

详细

Test #1:

score: 100
Accepted
time: 0ms
memory: 3584kb

input:

1
3 1 1
1 2 1
2 2 1

output:

1

result:

ok single line: '1'

Test #2:

score: -100
Wrong Answer
time: 99ms
memory: 3840kb

input:

1
1000 500 500
479 860 170
73 939 25
363 772 30
185 584 89
326 482 10
784 949 23
384 740 114
233 693 45
167 724 211
217 436 95
46 701 153
138 585 67
321 388 11
114 890 194
407 854 74
438 845 117
9 718 259
393 576 35
182 707 257
451 766 136
150 382 31
468 785 40
202 490 46
326 815 59
272 441 77
123 8...

output:

500

result:

wrong answer 1st lines differ - expected: '492', found: '500'