L2-013 红色警报 (25 分)(并查集)
额,说实话,还是第一次这样用并查集
本质就是记录连通项的个数。如果,轰炸一个城市,产生一个或者不产生,则说明,该城市对其他城市的连接没有直接作用。
那么怎么来数每次的连通项呢?没错!有多少个祖先就有多少个连通项。其实,就是每次都要把边连接起来,当然,被轰炸的城市所在的边用该丢弃!
#include<iostream> using namespace std; const int maxn = 5005; const int maxx = 505; struct node{ int u, v; }edge[maxn]; int fa[maxx]; int visit[maxx]; int n, m, u, v, k; int c; //并查集 int find(int x) { if (x == fa[x])return x; return fa[x] = find(fa[x]); } void unit(int u, int v){ int u1 = find(u); int v1 = find(v); fa[u1] = v1; } int main(){ cin >> n >> m; for (int i = 0; i < n; ++i)fa[i] = i; for (int i = 0; i < m; ++i) { cin >> u >> v; edge[i].u = u; edge[i].v = v; unit(u, v); } for (int i = 0; i < n; ++i)find(i); //记录原来的连通项数 int num1 = 0, num2; for (int i = 0; i < n; ++i)if (fa[i] == i)num1++; cin >> k; while(k--){ cin >> c; num2 = 0; visit[c] = 1; for (int i = 0; i < n; ++i)fa[i]=i; for (int i = 0; i < m; ++i){ if (visit[edge[i].u] == 1 || visit[edge[i].v] == 1)continue; else { unit(edge[i].u, edge[i].v); } } for (int i = 0; i < n; ++i)find(i); //记录现在的通项数 for (int i = 0; i < n; ++i)if (fa[i] == i)num2++; if (num2 == num1+1 || num2 == num1)cout << "City " << c << " is lost." << endl; else cout << "Red Alert: City " << c << " is lost!" << endl; num1 = num2; } num1 = 0; for (int i = 0; i < n;++i) if (visit[i] == 1)num1++; if (num1 == n)cout << "Game Over." << endl; }