题意:给出一个图,去除每条边的花费为边的长度,求用最少的花费去除部分边使得图中无圈。

思路:先将所有的边长加起来,然后减去最大生成树,即得出最小需要破坏的篱笆长度。

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std; int N, M; // 桩数量,篱笆数量
int par[10005];
void init() {
for (int i = 1; i <= N; ++i) par[i] = i;
}
int find(int x) {
return x == par[x] ? x : par[x] = find(par[x]);
}
bool same(int x, int y) {
return find(x) == find(y);
}
void unite(int x, int y) {
x = find(x);
y = find(y);
if (x != y) par[x] = y;
} struct point{
int x, y;
} ps[10005]; double dist(point &p1, point &p2) {
return sqrt((p1.x - p2.x)*(p1.x - p2.x) + (p1.y - p2.y)*(p1.y - p2.y));
} struct edge {
int from, to;
double cost;
edge(int from, int to, double cost) : from(from), to(to), cost(cost) {}
bool operator<(const edge &b) const { // 从大到小排序,求出最大生成树
return cost > b.cost;
}
}; vector<edge> es; // 边集
double ans = 0.0; // 答案 void kruskal() {
init();
sort(es.begin(), es.end());
for (auto it : es) { // C++11
if (!same(it.from, it.to)) {
unite(it.from, it.to);
ans -= it.cost; // 减去最大生成树的边即可
}
}
} void solve() {
kruskal();
printf("%.3lf\n", ans);
} int main()
{
cin >> N >> M;
for (int i = 1; i <= N; ++i)
cin >> ps[i].x >> ps[i].y;
int u, v;
double d;
for (int i = 0; i < M; ++i) {
cin >> u >> v;
d = dist(ps[u], ps[v]);
es.push_back(edge(u, v, d));
ans += d; // 求出所有路径和
}
solve();
return 0;
}
05-07 11:53