Fiasco

Fiasco

题意：给一个按bfs求最短路的代码（这个方法明显有bug），然后给出点和路段，问让你重新分配路径的值使得上诉错误的代码可以得到正确的值

解法：先把路径长度排序，再按bfs序从小到大分配

#include <iostream>
using namespace std;
#include <stdio.h>
#include <algorithm>
#include <cstring>

int a[30000][3], b[30000], map[2600][2600], tot, n, m, k;
bool flag[2600];

void sort(int i, int j) {
    int p = i, q = j;
    int mid = b[(p+q)/2];
    while (p <= q) {
        while (b[p] < mid) p++;
        while (b[q] > mid) q--;
        if (p <= q) {
            int temp = b[p];
            b[p] = b[q];
            b[q] = temp;
            p++; q--;
        }
    }
    if (i < q) sort(i, q);
    if (p < j) sort(p, j);
}

void bfs(int s) {
    int d[2600];
    d[1] = s;
    int i = 0, j = 1;
    while (i < j) {
        i++;
        flag[d[i]] = true;
        for (int k = 1; k <= n; k++) if (map[d[i]][k] != 0 && !flag[k]) {
            tot++;
            a[map[d[i]][k]][0] = b[tot];
            j++;
            d[j] = k;
            flag[k] = true;
        }
    }
}

int main() {
    int tt;
    cin >> tt;
    for (int cases = 1; cases <= tt; cases++) {
        cin >> n >> m >> k;
        memset(map, 0, sizeof(map));
        memset(flag, 0, sizeof(flag));
        for (int i = 1; i <= m; i++) {
            cin >> a[i][1] >> a[i][2];
            a[i][0] = 0;
            cin >> b[i];
            map[a[i][1]][a[i][2]] = i;
            map[a[i][2]][a[i][1]] = i;
        }
        sort(1, m);
        tot = 0;
        bfs(k);
        for (int i = 1; i <= m; i++) if (a[i][0] == 0) {
            tot++;
            a[i][0] = b[tot];
        }
        cout << "Case " << cases << ":" << endl;
        for (int i = 1; i <= m; i++)
            cout << a[i][1] << " " << a[i][2] << " "  << a[i][0] << endl;
    }
}