題目鏈接

題意

貓被困在圍欄里，問最少去掉多長的邊，使所有貓?zhí)映鰜怼?
問題轉(zhuǎn)化為將N個(gè)圖轉(zhuǎn)化為樹，因?yàn)闃洳粫森h(huán)，樹加上一條邊就可以成環(huán)

AC

• 并查集 + prim
  對于每個(gè)圖，求它的最大生成樹，總長度減去所有的最大生成樹就是需要去掉的邊

using namespace std; int inf = 0x3f3f3f3f; int n, m, pre[N]; double sum; bool vis[N]; double dis[N]; struct ac{int v;double c; }; vector<P> a(N); vector<ac> g[N]; void init() {for (int i = 1; i <N; ++i) {pre[i] = i;} } int find(int x) {if (x == pre[x]) return x;else return pre[x] = find(pre[x]); } void join(int x, int y) {int fx = find(x);int fy = find(y);if (fx == fy) return;else if (fx < fy) pre[fy] = fx;else pre[fx] = fy; }// 求最大生成樹 void prim(int x) {mem(dis, 0);mem(vis, false);for (int i = 0; i < g[x].size(); ++i) {ac t = g[x][i];dis[t.v] = t.c;}vis[x] = true;for (int i = 1; i < n; ++i) {double MAX = 0;int u = -1;for (int j = 1; j <= n; ++j) {if (vis[j]) continue;if (dis[j] > MAX) {MAX = dis[j];u = j;}}if (u == -1) return;vis[u] = true;sum += MAX;for (int j = 0; j < g[u].size(); ++j) {ac t = g[u][j];if (vis[t.v]) continue;if (dis[t.v] < t.c ) {dis[t.v] = t.c;}}} }int main(){ // #ifndef ONLINE_JUDGE // freopen("in.txt", "r", stdin); // #endifios::sync_with_stdio(false);while (cin >> n >> m) {init();for (int i = 1; i <= n; ++i) {cin >> a[i].first >> a[i].second;}double ans = 0;for (int i = 0; i < m; ++i) {int u, v;cin >> u >> v;int dx = a[u].first - a[v].first;int dy = a[u].second - a[v].second;double temp = sqrt(dx * dx + dy * dy);ans += temp;// 建圖g[u].push_back((ac){v, temp});g[v].push_back((ac){u, temp});join(u, v);}for (int i = 1; i <= n; ++i) {// 求這個(gè)并查集的最大生成樹if (pre[i] == i) {sum = 0;prim(i);ans -= sum;}}printf("%.3lf\n",ans);for (int i = 1; i <= n; ++i) {g[i].clear();}}return 0; }

• 貪心
  對所有邊降序排列，如果兩個(gè)邊不連通，就減去這條邊，剩下的就是需要去掉的邊

using namespace std; int inf = 0x3f3f3f3f; int n, m, pre[N]; struct ac{int u, v;double c;bool operator < (const ac &x)const {return c > x.c;} }; vector<P> a(N); vector<ac> g; void init() {for (int i = 1; i <N; ++i) {pre[i] = i;} } int find(int x) {if (x == pre[x]) return x;else return pre[x] = find(pre[x]); } bool join(int x, int y) {int fx = find(x);int fy = find(y);if (fx == fy) return false;else if (fx < fy) pre[fy] = fx;else pre[fx] = fy;return true; } double rm() {double sum = 0;for (int i = 0; i < g.size(); ++i) {ac t = g[i];if (join(t.u, t.v)) sum += t.c;}return sum; } int main(){ // #ifndef ONLINE_JUDGE // freopen("in.txt", "r", stdin); // #endifios::sync_with_stdio(false);while (cin >> n >> m) {init();for (int i = 1; i <= n; ++i) {cin >> a[i].first >> a[i].second;}double ans = 0;for (int i = 0; i < m; ++i) {int u, v;cin >> u >> v;int dx = a[u].first - a[v].first;int dy = a[u].second - a[v].second;double temp = sqrt(dx * dx + dy * dy);ans += temp;g.push_back((ac){u, v, temp});}sort(g.begin(), g.end());ans -= rm();printf("%.3lf\n",ans);g.clear();}return 0; }

總結(jié)

以上是生活随笔為你收集整理的Aizu 2224 Save your cats的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網(wǎng)站內(nèi)容還不錯(cuò)，歡迎將生活随笔推薦給好友。