Codeforces Round #212 (Div. 2) D. Fools and Foolproof Roads 并查集+优先队列
You must have heard all about the Foolland on your Geography lessons. Specifically, you must know that federal structure of this country has been the same for many centuries. The country consists of n cities, some pairs of cities are connected by bidirectional roads, each road is described by its length li.
The fools lived in their land joyfully, but a recent revolution changed the king. Now the king is Vasily the Bear. Vasily divided the country cities into regions, so that any two cities of the same region have a path along the roads between them and any two cities of different regions don't have such path. Then Vasily decided to upgrade the road network and construct exactly p new roads in the country. Constructing a road goes like this:
- We choose a pair of distinct cities u, v that will be connected by a new road (at that, it is possible that there already is a road between these cities).
- We define the length of the new road: if cities u, v belong to distinct regions, then the length is calculated as min(109, S + 1) (S — the total length of all roads that exist in the linked regions), otherwise we assume that the length equals 1000.
- We build a road of the specified length between the chosen cities. If the new road connects two distinct regions, after construction of the road these regions are combined into one new region.
Vasily wants the road constructing process to result in the country that consists exactly of q regions. Your task is to come up with such road constructing plan for Vasily that it meets the requirement and minimizes the total length of the built roads.
The first line contains four integers n (1 ≤ n ≤ 105), m (0 ≤ m ≤ 105), p (0 ≤ p ≤ 105), q (1 ≤ q ≤ n) — the number of cities in the Foolland, the number of existing roads, the number of roads that are planned to construct and the required number of regions.
Next m lines describe the roads that exist by the moment upgrading of the roads begun. Each of these lines contains three integers xi,yi, li: xi, yi — the numbers of the cities connected by this road (1 ≤ xi, yi ≤ n, xi ≠ yi), li — length of the road (1 ≤ li ≤ 109). Note that one pair of cities can be connected with multiple roads.
If constructing the roads in the required way is impossible, print a single string "NO" (without the quotes). Otherwise, in the first line print word "YES" (without the quotes), and in the next p lines print the road construction plan. Each line of the plan must consist of two distinct integers, giving the numbers of the cities connected by a road. The road must occur in the plan in the order they need to be constructed. If there are multiple optimal solutions, you can print any of them.
9 6 2 2
1 2 2
3 2 1
4 6 20
1 3 8
7 8 3
5 7 2
YES
9 5
1 9
Consider the first sample. Before the reform the Foolland consists of four regions. The first region includes cities 1, 2, 3, the second region has cities 4 and 6, the third region has cities 5, 7, 8, the fourth region has city 9. The total length of the roads in these cities is11, 20, 5 and 0, correspondingly. According to the plan, we first build the road of length 6 between cities 5 and 9, then the road of length 23 between cities 1 and 9. Thus, the total length of the built roads equals 29.
题意：
给你n点m边的无向图;
你可以加入p条任意边，而使得新图是由q个联通快构成的无向图
加边规则如下；
你可以选择两个不同点 相连，无论原来他们是否有边
你可以选择两个不同点相连，如果他们是不属于同一个联通快，那么新加入的边 的边权必须为 min(1e9,S+1)，S表示 这两个联通快的 总边权和
如果他们属于一个联通快，那么新加入的边 边权必须 为1000
相连之后，就属于一个联通快了
是否有方案构成q块
并且使得新加边的总边权最小
题解：
并查集维护联通快与边权和
优先队列每次选择联通快和最小的两个相连
最后多余的边都连在同样的两个点上就好了
#include<bits/stdc++.h> #include<queue> using namespace std; #define ls i<<1 #define rs ls | 1 #define mid ((ll+rr)>>1) #define pii pair<int,int> #define MP make_pair typedef long long LL; const long long INF = 1e18; const double Pi = acos(-1.0); const int N = 3e5+10, M = 2e5+35000+11, mod = 1e9+7, inf = 0x3fffffff; int n,m,p,q,edges[N],fa[N],num[N],a[N],vis[N]; vector<pii > ans; LL sum[N]; int finds(int x) {return fa[x] == x? x:fa[x]=finds(fa[x]);} struct node{LL value;int id; bool operator < (const node &r) const { return value > r.value; } }; int main() { scanf("%d%d%d%d",&n,&m,&p,&q); for(int i = 1; i <= n; ++i) fa[i] = i,sum[i] = 0, num[i] = 1; for(int i = 1; i <= m; ++i) { int u,v,w; scanf("%d%d%d",&u,&v,&w); int fx = finds(u); int fy = finds(v); sum[fx] += w; if(fx!=fy) { num[fx] += num[fy]; fa[fy] = fx; sum[fx] += sum[fy]; } } priority_queue<node> Q; int block = 0; for(int i = 1; i <= n; ++i) { int fx = finds(i); if(!vis[fx]) { Q.push(node{sum[fx],fx}); // cout<<sum[fx]<<" "<<fx<<endl; block++; vis[fx] = 1; } } block = block - q; if(block < 0) { puts("NO"); return 0; } while(!Q.empty() && block--) { node k = Q.top(); Q.pop(); if(Q.empty()) {break;} node k2 = Q.top(); Q.pop(); // cout<<k.id<<" "<<k2.id<<endl; ans.push_back(MP(k.id,k2.id)); num[k.id] += num[k2.id]; fa[k2.id] = fa[k.id]; Q.push(node{k.value+k2.value+min(1000000000LL,k.value+k2.value+1),k.id}); p--; } if(p < 0) { puts("NO"); return 0; } if(p) { int flag = -1; for(int i = 1; i <= n; ++i) { int fx = finds(i); if(num[fx]>1) { flag = fx; // cout<<fx<<endl; break; } } for(int cnt = 0,i = 1; i <= n; ++i) { if(finds(i) == flag) { a[++cnt] = i; } if(cnt == 2) break; } if(flag == -1) { puts("NO");return 0; } while(p--) { ans.push_back(MP(a[1],a[2])); } } puts("YES"); for(int i = 0; i < ans.size(); ++i) cout<<ans[i].first<<" "<<ans[i].second<<endl; return 0; }
