（最小生成树）Codeforces 76 A Gift

Posted 2020-09-20 惜取少年时

The kingdom of Olympia consists of N cities and M bidirectional roads. Each road connects exactly two cities and two cities can be connected with more than one road. Also it possible that some roads connect city with itself making a loop.

All roads are constantly plundered with bandits. After a while bandits became bored of wasting time in road robberies, so they suggested the king of Olympia to pay off. According to the offer, bandits want to get a gift consisted of gold and silver coins. Offer also contains a list of restrictions: for each road it is known g_i — the smallest amount of gold and s_i — the smallest amount of silver coins that should be in the gift to stop robberies on the road. That is, if the gift contains a gold and b silver coins, then bandits will stop robberies on all the roads that g_i?≤?a and s_i?≤?b.

Unfortunately kingdom treasury doesn‘t contain neither gold nor silver coins, but there are Olympian tugriks in it. The cost of one gold coin in tugriks is G, and the cost of one silver coin in tugriks is S. King really wants to send bandits such gift that for any two cities there will exist a safe path between them. Your task is to find the minimal cost in Olympian tugriks of the required gift.

Input

The first line of the input contains two integers N and M (2?≤?N?≤?200, 1?≤?M?≤?50?000) — the number of cities and the number of roads, respectively. The second line contains two integers G and S (1?≤?G,?S?≤?10⁹) — the prices of gold and silver coins in tugriks. The following M lines contain information about the offer. Each of the records in list is given as four integers x_i,?y_i,?g_i,?s_i, where x_i and y_i are the numbers of cities that the road connects and g_i, s_i are minimal gold and silver coins requirements for the i-th road (1?≤?x_i,?y_i?≤?N, 1?≤?g_i,?s_i?≤?10⁹). Cities are numbered from 1 to N. It is possible that there are more than one road between a pair of cities. It is possible that a road connects the city with itself.

Output

The output should contain the minimal cost of the gift in Olympian tugriks. If there is no gift that satisfies the given requirements output .

Example

Input

3 3
2 1
1 2 10 15
1 2 4 20
1 3 5 1

Output

30

 

调整g值观察，注意到随着g的单调增加，s逐渐变小。故采取枚举g的值，依次建立最小生成树比较的做法，并且注意到每次增加g时只会多出一个边，即每次只可能由之前的生成树与新加的一条边来共同组成新的生成树，这样就减少了sort的过程，将新的边插入到合适的位置即可，复杂度降低了一个logm。

  1 #include <iostream>
  2 #include <string>
  3 #include <algorithm>
  4 #include <cstring>
  5 #include <cstdio>
  6 #include <cmath>
  7 #include <queue>
  8 #include <set>
  9 #include <map>
 10 #include <list>
 11 #include <vector>
 12 #include <stack>
 13 #define mp make_pair
 14 #define MIN(a,b) (a>b?b:a)
 15 #define rank rankk
 16 //#define MAX(a,b) (a>b?a:b)
 17 typedef long long ll;
 18 typedef unsigned long long ull;
 19 const int MAX=5e4+5;
 20 const ll INF=9223372036854775807;
 21 const int B=1024;//桶的大小
 22 const double M=4e18;
 23 using namespace std;
 24 const int MOD=1e9+7;
 25 typedef pair<int,int> pii;
 26 const double eps=0.000000001;
 27 ll maxg,maxs;
 28 int num;
 29 int n,m;
 30 /*
 31     kruskal最小生成树算法
 32     按照边的权值的顺序从小到大看一遍，如果不产生圈（重边也考虑在内）
 33     就把当前这条边加入到生成树中
 34     是否产生圈只需看两点之前是否在同一连通分量里
 35     使用并查集判断是否属于同一个连通分量
 36 */
 37 #define rank rankk  //由于与某个库名称相同，故事先define
 38 /*
 39 并查集
 40 复杂度为阿克曼函数的反函数，比O(log(n))还快
 41 */
 42 
 43 int par[MAX];//父亲
 44 int rank[MAX];//树的高度
 45 //初始化n个元素
 46 void init(int n)
 47 {
 48     for(int i=0;i<n;i++)
 49     {
 50         par[i]=i;
 51         rank[i]=0;
 52     }
 53 }
 54 //查询树的根，期间加入了路径压缩
 55 int find(int x)
 56 {
 57     if(par[x]==x)
 58         return x;
 59     else
 60         return par[x]=find(par[x]);
 61 }
 62 //合并x和y所属的集合
 63 void unite(int x,int y)
 64 {
 65     x=find(x);
 66     y=find(y);
 67     if(x==y)
 68         return ;
 69     if(rank[x]<rank[y])
 70         par[x]=y;
 71     else
 72     {
 73         par[y]=x;
 74         if(rank[x]==rank[y])
 75             rank[x]++;
 76     }
 77 }
 78 //判断x和y是否属于同一个集合
 79 bool same(int x,int y)
 80 {
 81     return find(x)==find(y);
 82 }
 83 /*
 84 建立结构体记录边
 85 */
 86 struct edge
 87 {
 88     int u,v;
 89     ll cost,cost2;
 90 };
 91 bool cmp(const edge &e1,const edge &e2)
 92 {
 93     if(e1.cost!=e2.cost)
 94         return e1.cost<e2.cost;
 95     else
 96         return e1.cost2<e2.cost2;
 97 }
 98 bool cmp2(const edge &e1,const edge &e2)
 99 {
100     if(e1.cost2!=e2.cost2)
101         return e1.cost2<e2.cost2;
102     else
103         return e1.cost<e2.cost;
104 }
105 bool kruskal(edge *e,int vertice_num)//e为存储边的数组，下标从0开始
106 {
107     maxg=maxs=0LL;
108     init(n+2);//初始化并查集
109     num=0;
110     for(int i=1;i<=vertice_num;i++)
111     {
112         edge tem=e[i];
113         if(!same(tem.u,tem.v))
114         {
115             unite(tem.u,tem.v);
116             e[++num]=e[i];
117             maxg=max(maxg,e[i].cost);
118             maxs=max(maxs,e[i].cost2);
119             if(num==n-1)
120                 break;
121         }
122     }
123     if(num==n-1)
124         return true;
125     else
126         return false;
127 }
128 edge es[MAX];
129 ll g,s,an;
130 int main()
131 {
132     while(~scanf("%d%d",&n,&m))
133     {
134         an=INF;
135 //        scanf("%lld%lld",&g,&s);
136         scanf("%I64d%I64d",&g,&s);
137         for(int i=1;i<=m;i++)
138         {
139 //            scanf("%d%d%lld%lld",&es[i].u,&es[i].v,&es[i].cost,&es[i].cost2);
140             scanf("%d%d%I64d%I64d",&es[i].u,&es[i].v,&es[i].cost,&es[i].cost2);
141         }
142         sort(es+1,es+1+m,cmp);
143 //        sort(es+1,es+1+n-1,cmp2);
144 //        if(kruskal(es,n-1))
145 //            an=min(an,maxg*g+maxs*s);
146 
147         for(int i=1;i<=m;i++)
148         {
149 //            es[i<=n?i:n]=es[i];
150 //            int j=(i<=n?i-1:n-1);
151             es[++num]=es[i];
152             int j=num-1;
153             while(j&&es[j].cost2>es[j+1].cost2)
154             {
155                 swap(es[j],es[j+1]);
156                 --j;
157             }
158             if(kruskal(es,num))
159                 an=min(an,maxg*g+maxs*s);
160         }
161         if(an==INF)
162             printf("-1\n");
163         else
164 //            printf("%lld\n",an);
165             printf("%I64d\n",an);
166     }
167 
168 }