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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
#203028#2483. Roof EscapeSolitaryDream#RE 0ms0kbC++171.7kb2023-10-06 14:48:072023-10-06 14:48:08

Judging History

你现在查看的是最新测评结果

  • [2023-10-06 14:48:08]
  • 评测
  • 测评结果:RE
  • 用时:0ms
  • 内存:0kb
  • [2023-10-06 14:48:07]
  • 提交

answer

#include <bits/stdc++.h>
using namespace std;
#define int long long
const int N = 1e5 + 10;
int n, m, sx, sy, ex, ey;
vector<int> mp[N];
inline double Sqr(double x) {
    return x * x;
}
inline int Det(int a, int b, int u, int v, int x, int y) {
    u -= a; v -= b;
    x -= a; y -= b;
    return u * y - x * v;
}
signed main() {
    scanf("%lld%lld%lld%lld%lld%lld", &n, &m, &sx, &sy, &ex, &ey);
    n /= 2; m /= 2;
    for (int i = 1; i <= n; ++i) {
        mp[i].resize(m + 2);
        for (int j = 1; j <= m; ++j) scanf("%lld", &mp[i][j]);
    }
    if (sx > ex) {
        sx = 2 * n - sx; ex = 2 * n - ex;
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j) 
                if (i < n - i + 1) swap(mp[i][j], mp[n - i + 1][j]);
    }
    if (sy > ey) {
        sy = 2 * m - sy; ey = 2 * m - ey;
        for (int i = 1; i <= n; ++i)
            for (int j = 1; j <= m; ++j) 
                if (j < m - j + 1) swap(mp[i][j], mp[i][m - j + 1]);
    }
    double ans1 = sqrt(Sqr(sx - ex) + Sqr(sy - ey));
    int ans2 = 0;
    for (int i = (sx + 1) / 2, j = (sy + 1) / 2; ; ) {
        int d = Det(sx, sy, 2 * i + 2, 2 * j + 2, ex, ey);
        if (d > 0) {
            ans2 += abs(mp[i][j] - mp[i][j + 1]);
            ++j;
        } else if (d == 0) {
            vector<int> a({mp[i][j], mp[i + 1][j], mp[i][j + 1], mp[i + 1][j + 1]});
            sort(a.begin(), a.end());
            int h = a[2];
            ans2 += abs(mp[i][j] - h) + abs(mp[i + 1][j + 1] - h);
            ++i; ++j;
        } else {
            ans2 += abs(mp[i][j] - mp[i + 1][j]);
            ++i;
        }
        if (i == (ex + 1) / 2 && j == (ey + 1) / 2) break; 
    }    
    printf("%.10lf\n", ans1 + ans2);
    return 0;
}

詳細信息

Test #1:

score: 0
Runtime Error

input:

4 26 1 1 3 25
0 1
0 5
5 5
0 1
2 1
4 1
5 0
0 0
1 0
6 4
3 2
5 4
1 5

output:


result: