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HDU4085 Peach Blossom Spring

2024-09-01 21:32:28   218人阅读

Tao Yuanming(365-427) was a Chinese poet of Eastern Jin dynasty. One of his most famous works is "Peach Blossom Spring", which is a fable about a chance 
discovery of an ethereal village where the people lead an ideal existence in harmony with nature, unaware of the outside world for centuries. So in Chinese, "Peach Blossom Spring" means "utopia". 
In the story of "Peach Blossom Spring", there was a mysterious place. In Qin dynasty, some people escaped to that place during the civil unrest and built a village. They and their descendants never left and never had any contact with the outside world since then, until centuries latter a fisherman of Jin dynasty found them. 
Recently, some Chinese ACMers happened to find the relics of the village mentioned in"Peach Blossom Spring". 
They also found a document about building hiding places to escape from Qin army. The document said: 
There were n houses and m roads in the village. Each road connected two houses. These houses were numbered from 1 to n. There were k families, each living in a different house. 
The houses they lived were house 1, house 2, … , house k. There were also k broken houses: house n-k+1, house n-k+2, ... , house n, with secret basements so that those houses could be used as hiding places. 
The problem was that all roads were broken. People wanted to repair some roads so that every family could reach a hiding place through the repaired roads. Every hiding place could only hold one family. Each road cost some labor to be repaired. The head of the village wanted to find out the minimum cost way of repairing the roads, but he didn‘t know how to do. 
Would you solve the problem which the ancient village head never solved?

InputThe input begins with a line containing an integer T(T<=50), the number of test cases. For each case, the first line begins with three integers ---- the above mentioned n (4<=n<=50), m (0<=m<=1000) and k (1<=k<=5, 2k<=n). Then m lines follow, each containing three integers u,v and w, indicating that there is a broken road connecting house u an d v, and the cost to repair that road is w(1<=w<=1000).OutputFor each test case, if you cannot find a proper way to repair the roads, output a string "No solution" in a line. Otherwise, output the minimum cost to repair the roads in a line.Sample Input

24 3 14 2 103 1 92 3 106 7 21 5 10002 6 10001 3 12 3 13 4 14 5 14 6 1

Sample Output

295

斯坦纳树

求包含前k个点和后k个点的最小生成子树。

用状态压缩来表示某个点是否被包含进了子树，然后跑一大堆SPFA。

按照题目的要求，前k个点和后k个点只要一对在一棵子树中即可，不需要所有2*k个点在同一棵子树中，所以最后要在森林中DP。

并不明白SPFA的初始状态为什么要那么写，注释掉的代码wa掉了。先抄题解为敬2333

 

  1 /*by SilverN*/  2 #include<iostream>  3 #include<algorithm>  4 #include<cstring>  5 #include<cstdio>  6 #include<cmath>  7 #include<queue>  8 using namespace std;  9 const int INF=0x3f3f3f3f; 10 const int mxn=2100; 11 int read(){ 12     int x=0,f=1;char ch=getchar(); 13     while(ch<‘0‘ || ch>‘9‘){if(ch==‘-‘)f=-1;ch=getchar();} 14     while(ch>=‘0‘ && ch<=‘9‘){x=x*10+ch-‘0‘;ch=getchar();} 15     return x*f; 16 } 17 struct edge{ 18     int v,nxt,dis; 19 }e[mxn]; 20 int hd[mxn],mct=0; 21 void add_edge(int u,int v,int d){ 22     e[++mct].v=v;e[mct].nxt=hd[u];e[mct].dis=d;hd[u]=mct;return; 23 } 24 int n,m,k; 25 int id[60]; 26 int dis[mxn][60]; 27 bool vis[mxn][60]; 28 bool inq[mxn][60]; 29 void init(){ 30     memset(hd,0,sizeof hd); 31     memset(vis,0,sizeof vis); 32     memset(dis,0x3f,sizeof dis); 33     memset(id,0,sizeof id); 34     mct=0; 35     return; 36 } 37 queue<pair<int,int> >q; 38 void SPFA(){ 39     while(!q.empty()){ 40         int w=q.front().first; 41         int u=q.front().second; 42         q.pop(); 43         vis[w][u]=0; 44         for(int i=hd[u];i;i=e[i].nxt){ 45             int v=e[i].v; 46             if(dis[w|id[v]][v]>dis[w][u]+e[i].dis){ 47                 dis[w|id[v]][v]=dis[w][u]+e[i].dis; 48                 if(!vis[v][w]){ 49                     q.push(make_pair(w|id[v],v)); 50                     vis[w|id[v]][v]=1; 51                 } 52             } 53         } 54     } 55     return; 56 } 57 bool ok(int w){ 58     int res=0; 59     for(int i=1;w;i++){ 60         if(w&1){ 61             if(i>k)res++;//保持出入数量相等  62             else res--; 63         } 64         w>>=1; 65     } 66     if(res)return 0;return 1; 67 } 68 int dp[mxn]; 69 void DP(){ 70     int i,j;int ed=1<<(2*k); 71     memset(dp,0x3f,sizeof dp); 72     for(i=0;i<ed;i++) 73         for(j=1;j<=n;j++) 74             dp[i]=min(dp[i],dis[i][j]); 75     for(i=1;i<ed;i++){ 76         if(ok(i)){ 77             for(j=(i-1)&i;j;j=(j-1)&i){ 78                 if(ok(j)){ 79                     dp[i]=min(dp[i],dp[j]+dp[i-j]);//合并状态  80                 } 81             } 82         } 83     } 84     return; 85 } 86 int main(){ 87     int i,j; 88     int T=read(); 89     while(T--){ 90         init(); 91         n=read();m=read();k=read(); 92         int u,v,d; 93         for(i=1;i<=m;i++){ 94             u=read();v=read();d=read(); 95             add_edge(u,v,d); 96             add_edge(v,u,d); 97         } 98         for(i=1;i<=k;i++){ 99             id[i]=1<<(i-1);100             id[n-i+1]=1<<(k+i-1);101             dis[id[i]][i]=0;102             dis[id[n-i+1]][n-i+1]=0;103         }104         int ed=1<<(2*k);105         for(int s=0;s<ed;s++){106             for(int i=1;i<=n;i++){107                 for(int p=(s-1)&s;p;p=(p-1)&s)108                     dis[s][i]=min(dis[s][i],dis[p|id[i]][i]+dis[(s-p)|id[i]][i]);109                 if(dis[s][i]<INF && !vis[s][i])110                     q.push(make_pair(s,i));vis[s][i]=1;111             }112             SPFA();113         }114 /*        for(i=1;i<=n;i++){115             if(id[i]){116                 vis[id[i]][i]=1;117                 q.push(make_pair(id[i],i));118             }119         }120         SPFA();*/121         DP();122         if(dp[ed-1]>=INF)printf("No solution\n");123         else printf("%d\n",dp[ed-1]);124     }125     return 0;126 }

 

HDU4085 Peach Blossom Spring


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