最小生成树-字典序最小

最小生成树-字典序最小

求满足字典序最小的最小生成树，并输出。

#include<iostream>#include<algorithm>#include<stdio.h>#include<cstring>#include<string>using namespace std;const int MAXN = 110;struct Node{int u;int v;int w;};Node E[MAXN*MAXN];//保存边集Node R[MAXN*MAXN];//保存字典序最小路径int map[MAXN][MAXN], n, vset[MAXN], ans;//保存各点关系的举证，顶点数，辅助数组【是否在同一个集合】，路径数bool cmp(Node a, Node b)//贪心，按权值最小{if (a.w == b.w){if (a.u == b.u){return a.v < b.v;}return a.u < b.u;}return a.w < b.w;}bool cmp_road(Node a, Node b){if (a.u == b.u){return a.v < b.v;}return a.u < b.u;}void printf_road(){if (ans != n - 1)//n个顶点只需n-1条边{cout << "该图非连通图" << endl;return;}sort(R, R + ans, cmp_road);//字典序最小排序for (int i = 0; i < ans; i++){cout << "from" << R[i].u << "to" << R[i].v << " value:" << R[i].w << endl;}}int find(int x)//判断2个顶点是在同一个集合{if (vset[x] == -1){return x;}return vset[x] = find(vset[x]);}void kruskal(){int k = 0;for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){if (map[i][j]!=0)//map[i][j]==0等于无边{E[k].u = i;E[k].v = j;E[k].w = map[i][j];k++;}}}ans = 0;sort(E, E + k, cmp);memset(vset, -1, sizeof(vset));for (int j = 0; j < k&&ans<n-1; j++){int u = E[j].u;int v = E[j].v;int t1 = find(u);int t2 = find(v);if (t1 != t2){R[ans++] = E[j];vset[t1] = t2;}}printf_road();}int main(){cin >> n;for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){cin >> map[i][j];}}kruskal();return 0;}

例题：ZOJ 3204

Connect them

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Time Limit: 1 Second      Memory Limit: 32768 KB

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You have n computers numbered from 1 to n and you want to connect them to make a small local area network (LAN). All connections are two-way (that is connecting computers i and j is the same as connecting computers j and i). The cost of connecting computer i and computer j is c_ij. You cannot connect some pairs of computers due to some particular reasons. You want to connect them so that every computer connects to any other one directly or indirectly and you also want to pay as little as possible.

Given n and each c_ij , find the cheapest way to connect computers.

Input

There are multiple test cases. The first line of input contains an integer T (T <= 100), indicating the number of test cases. Then T test cases follow.

The first line of each test case contains an integer n (1 < n <= 100). Then n lines follow, each of which contains n integers separated by a space. The j-th integer of the i-th line in these n lines is c_ij, indicating the cost of connecting computers i and j (c_ij = 0 means that you cannot connect them). 0 <= c_ij <= 60000, c_ij = c_ji, c_ii = 0, 1 <= i, j <= n.

Output

For each test case, if you can connect the computers together, output the method in in the following fomat:

i₁ j₁ i₁ j₁ ......

where i_k i_k (k >= 1) are the identification numbers of the two computers to be connected. All the integers must be separated by a space and there must be no extra space at the end of the line. If there are multiple solutions, output the lexicographically smallest one (see hints for the definition of "lexicography small") If you cannot connect them, just output "-1" in the line.

Sample Input

230 2 32 0 53 5 020 00 0

Sample Output

1 2 1 3-1

Hints:
A solution A is a line of p integers: a₁, a₂, ...a_p.
Another solution B different from A is a line of q integers: b₁, b₂, ...b_q.
A is lexicographically smaller than B if and only if:
(1) there exists a positive integer r (r <= p, r <= q) such that a_i = b_i for all 0 < i < r and a_r < b_r 
OR
(2) p < q and a_i = b_i for all 0 < i <= p

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Author: CAO, Peng
Source: The 6th Zhejiang Provincial Collegiate Programming Contest

#include<iostream>#include<algorithm>#include<stdio.h>#include<cstring>#include<string>using namespace std;const int MAXN = 110;struct Node{int u;int v;int w;};Node E[MAXN*MAXN];struct road{int u, v;};road R[MAXN*MAXN];int map[MAXN][MAXN], n, vset[MAXN], ans;bool cmp(Node a, Node b){if (a.w == b.w){if (a.u == b.u){return a.v < b.v;}return a.u < b.u;}return a.w < b.w;}bool cmp_road(road a, road b){if (a.u == b.u){return a.v < b.v;}return a.u < b.u;}void printf_road(){sort(R, R + ans, cmp_road);for (int i = 0; i < ans; i++){cout << R[i].u << ' ' << R[i].v;if (i != ans - 1){cout << ' ';}}cout << endl;}void kruskal(){int k = 0;for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){if (j>i&&map[i][j]){E[k].u = i;E[k].v = j;E[k].w = map[i][j];k++;}}}ans = 0;sort(E, E + k, cmp);for (int i = 1; i <= n; i++){vset[i] = i;}for (int j = 0; j < k; j++){int t1 = vset[E[j].u];int t2 = vset[E[j].v];if (t1 != t2){R[ans].u = E[j].u;R[ans++].v = E[j].v;for (int i = 1; i <= n; i++){if (vset[i] == t2){vset[i] = t1;}}}}}int main(){int t;cin >> t;while (t--){cin >> n;for (int i = 1; i <= n; i++){for (int j = 1; j <= n; j++){cin >> map[i][j];}}kruskal();if (ans != n - 1){cout << "-1" << endl;}else{printf_road();}}return 0;}