题解 | #最短路径#

//先用Dijkstra硬写了一遍
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <typeindex>
#include <vector>
#include <cmath>
using namespace std;

int list2[1000][1000] = {0}, n, m;

typedef struct node {
    int a;
    int b;
} ;

int pow1(int j) {
    int all = 1;
    for (int i = 0; i < j; i++)
        all = all * 2 % 100000;
    return all;
}

void add(int a, int b, int o) {
    for (int i = 0; i < m; i++)
        list2[a][i] = list2[b][i];
    list2[a][o] = 1;
    return;
}

bool compare(int a, int o, int b, int j) {
    for (int i = m - 1; i >= 0; i--) {
        int i1 = (i == o) ? list2[a][i] + 1 : list2[a][i];
        int i2 = (i == j) ? list2[b][i] + 1 : list2[b][i];
        if (i1 > i2 )
            return true;
        if (i1 < i2)
            return false;
    }
    return false;
}

int main() {
    while (cin >> n >> m) { // 注意 while 处理多个 case
        vector<node>list(m);
        int list1[n];
        int list3[n];
        memset(list1, 0, sizeof(list1));
        memset(list3, 0, sizeof(list1));
        list1[0] = 1;
        for (int i = 0; i < m; i++)
            cin >> list[i].a >> list[i].b;
        for (int i = 0; i < n; i++) {
            int p = 0, q = 0, o = 0, max = 0, k;
            for (int j = 0; j < m; j++)
                if (list1[list[j].a]  + list1[list[j].b] == 1 ) {
                    int g = list1[list[j].a] == 1 ? list[j].a : list[j].b;
                    if (compare(p, o, g,  j) || q == 0) {
                        p = g;
                        q = list1[list[j].a] == 1 ? list[j].b : list[j].a;
                        o = j;
                    }
                }
            add(q, p, o) ;
            list1[q] = 1, max = k;
            list3[q] = (list3[p] + pow1(o)) % 100000;
        }

        for (int j = 1; j < n; j++)
            cout << list3[j] << endl;
    }
}
// 64 位输出请用 printf("%lld")

//后来发现一些代码删掉也可以得到正确结果，看了其他大佬的答案，发现这道题用Prim也可以



#include <cstdlib>
#include <cstring>
#include <iostream>
#include <typeindex>
#include <vector>
#include <cmath>
using namespace std;

int list2[100][500] = {0}, n, m;

int pow1(int j) {
    int all = 1;
    for (int i = 0; i < j; i++)
        all = all * 2 % 100000;
    return all;
}

void add(int a, int b, int o) {
    for (int i = 0; i < m; i++)
        list2[a][i] = list2[b][i];
    list2[a][o] = 1;
    return;
}

int main() {
    while (cin >> n >> m) { // 注意 while 处理多个 case
        int list[500][2] = {0}, list1[100] = {0}, list3[100] = {0}, p, q;
        list1[0] = 1;
        for (int i = 0; i < m; i++)
            cin >> list[i][0] >> list[i][1];

        for (int i = 0; i < n; i++)
            for (int j = 0; j < m; j++)
                if (list1[list[j][0]]  + list1[list[j][1]] == 1 ) {
                    p = list1[list[j][0]] == 1 ? list[j][0] : list[j][1];
                    q = list1[list[j][0]] == 1 ? list[j][1] : list[j][0];
                    add(q, p, j) ;
                    list1[q] = 1, list3[q] = (list3[p] + pow1(j)) % 100000;
                    break;
                }

        for (int j = 1; j < n; j++)
            cout << list3[j] << endl;
    }
}
// 64 位输出请用 printf("%lld")