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ACM Contest and Blackout
题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=1541
In order to prepare the “The First National ACM School Contest” (in 20??) the major of the city decided to provide all the schools with a reliable source of power. (The major is really afraid of blackoutsJ). So, in order to do that, power station “Future” and one school (doesn’t matter which one) must be connected; in addition, some schools must be connected as well.
You may assume that a school has a reliable source of power if it’s connected directly to “Future”, or to any other school that has a reliable source of power. You are given the cost of connection between some schools. The major has decided to pick out two the cheapest connection plans – the cost of the connection is equal to the sum of the connections between the schools. Your task is to help the major — find the cost of the two cheapest connection plans.
Input
The Input starts with the number of test cases, T (1 < T < 15) on a line. Then T test cases follow. The first line of every test case contains two numbers, which are separated by a space, N (3 < N < 100) the number of schools in the city, and M the number of possible connections among them. Next M lines contain three numbers Ai , Bi , Ci , where Ci is the cost of the connection (1 < Ci < 300) between schools Ai and Bi . The schools are numbered with integers in the range 1 to N.
Output
For every test case print only one line of output. This line should contain two numbers separated by a single space – the cost of two the cheapest connection plans. Let S1 be the cheapest cost and S2 the next cheapest cost. It’s important, that S1 = S2 if and only if there are two cheapest plans, otherwise S1 < S2. You can assume that it is always possible to find the costs S1 and S2.
Sample Input
2
5 8
1 3 75
3 4 51
2 4 19
3 2 95
2 5 42
5 4 31
1 2 9
3 5 66
9 14
1 2 4
1 8 8
2 8 11
3 2 8
8 9 7
8 7 1
7 9 6
9 3 2
3 4 7
3 6 4
7 6 2
4 6 14
4 5 9
5 6 10
Sample Output
110 121
37 37
题意：求最小生成树和次小生成树的权值。注意次小生成树和最小生成树权值可能相同。
思路：次小生成树模板题。用二维数组maxd优化，求出次小生成树。
AC代码：
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<queue>
#include<string>
#include<vector>
using namespace std;
typedef long long ll;
const int INF = 0x3f3f3f3f;
const int MAXN = 1e5 + 7;
int n, m;
struct edge {
int u, v;
int cost, used;
}e[MAXN];
int pre[MAXN];
int maxd[1010][1010];
vector<int>G[1010];
bool cmp(edge a, edge b);
int find(int x);
void Union(int a, int b);
int main() {
int t;
scanf("%d", &t);
while (t--) {
int sum = 0;
scanf("%d%d", &n, &m);
memset(maxd, 0, sizeof(maxd));
for (int i = 0; i <= n; ++i) {
pre[i] = i;
G[i].clear();
G[i].push_back(i);
}
for (int i = 1; i <= m; ++i) {
scanf("%d%d%d", &e[i].u, &e[i].v, &e[i].cost);
e[i].used = 0;
}
sort(e + 1, e + 1 + m, cmp);
for (int i = 1; i <= m; ++i) {
int u = find(e[i].u), v = find(e[i].v);
int cost = e[i].cost;
if (u != v) {
sum += e[i].cost;
e[i].used = 1;
for (int k = 0; k<G[u].size(); ++k) {
for (int j = 0; j<G[v].size(); ++j) {
maxd[G[u][k]][G[v][j]] = maxd[G[v][j]][G[u][k]] = cost;
}
}
Union(u, v);
for (int k = 0; k<G[u].size(); ++k) {
G[v].push_back(G[u][k]);
}
}
}
int ans = INF;
for (int i = 1; i <= m; ++i) {
if (!e[i].used) {
ans = min(ans, sum + e[i].cost - maxd[e[i].u][e[i].v]);
}
}
printf("%d %d\n", sum, ans);
}
return 0;
}
bool cmp(edge a, edge b) {
return a.cost<b.cost;
}
int find(int x) {
if (pre[x] == x)
return x;
return pre[x] = find(pre[x]);
}
void Union(int a, int b) {
int x = find(a);
int y = find(b);
if (x == y)
return;
pre[x] = y;
}相似题型：UVA - 10462