Picnic Planning POJ - 1639最小限度生成树

5000+B，调死我了

set里判重居然是用符号比较判的，调出这个问题之后又莫名re，maxn改到3000居然才能跑discuss的样例

后来本机测发现弹了个bad_alloc。。看来以后做题vector不能随便开这么多了，估计改成静态数组每个一个tot就可以maxn30了

  1 //#include<bits/stdc++.h>  
  2 //#pragma comment(linker, "/STACK:1024000000,1024000000")   
  3 #include<stdio.h>  
  4 #include<algorithm>  
  5 #include<queue>  
  6 #include<string.h>  
  7 #include<iostream>  
  8 #include<math.h>                    
  9 #include<stack>
 10 #include<set>  
 11 #include<map>  
 12 #include<vector>  
 13 #include<iomanip> 
 14 #include<bitset>
 15 using namespace std;         //
 16 
 17 #define ll long long  
 18 #define ull unsigned long long
 19 #define pb push_back  
 20 #define FOR(a) for(int i=1;i<=a;i++) 
 21 #define sqr(a) (a)*(a)
 22 #define dis(a,b) sqrt(sqr(a.x-b.x)+sqr(a.y-b.y))
 23 ll qp(ll a,ll b,ll mod){
 24     ll t=1;while(b){if(b&1)t=t*a%mod;b>>=1;a=a*a%mod;}return t;
 25 }
 26 struct DOT{int x;int y;};
 27 inline void read(int &x){int k=0;char f=1;char c=getchar();for(;!isdigit(c);c=getchar())if(c=='-')f=-1;for(;isdigit(c);c=getchar())k=k*10+c-'0';x=k*f;} 
 28 void ex(){puts("No");exit(0);}
 29 const int dx[4]={0,0,-1,1};
 30 const int dy[4]={1,-1,0,0};
 31 const int inf=0x3f3f3f3f; 
 32 const ll Linf=0x3f3f3f3f3f3f3f3fLL;
 33 const ll Mod=1e18+7;
 34 
 35 const int maxn=3300;
 36 
 37 int n,lim;
 38 map<string,int>idx;
 39 char buf[2][15];
 40 int tot;
 41 int edgcnt;
 42 
 43 struct EDGE{
 44     int u,v,w;
 45     bool del;
 46     int id;
 47     friend bool operator <(EDGE A,EDGE B){return A.w<B.w;}
 48 };
 49 set<int>DES;
 50 vector<EDGE>G[maxn];
 51 vector<EDGE>T[maxn];    //树图
 52 
 53 int belong[maxn];
 54 vector<int>hav[maxn];
 55 int blkcnt;
 56 int root;
 57 
 58 int fa[maxn];
 59 void init(int n){for(int i=1;i<=n;i++)fa[i]=i;}
 60 int find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}
 61 void unite(int x,int y){x=find(x);y=find(y);fa[x]=y;}
 62 
 63 vector<EDGE>BLK;
 64 
 65 void dfs(int x,int fa){
 66     if(belong[x])return;
 67     belong[x]=blkcnt;
 68     hav[blkcnt].pb(x);
 69     for(int i=0;i<G[x].size();i++){
 70         int v=G[x][i].v,w=G[x][i].w;
 71         if(v==fa)continue;
 72         if(v==root)continue;
 73         dfs(v,x);
 74     }
 75 }
 76 
 77 
 78 int ans;
 79 
 80 set<int>SSS;    //park连接的点
 81 int du;            //root的度数
 82 
 83 struct NODE{
 84     EDGE del;
 85     int val;
 86     int to;
 87     int newLen;    //
 88     friend bool operator<(NODE A,NODE B){
 89         return A.val<B.val;
 90     }
 91 };
 92 
 93 int flag;
 94 
 95 void dfsx(int u,int fa,int &mxlen,EDGE &remo,int gf){
 96     //if(ans==82 && gf==6){
 97     //    cout<<u<<"路径"<<T[6].size()<<endl;
 98     //}
 99     if(u==root){
100         flag=1;return;
101     }
102 
103 
104     for(int i=0;i<T[u].size();i++){
105 
106         if(DES.count(T[u][i].id))continue;
107 
108         //if(ans==82 && gf==6 && u==6){
109         //    cout<<"goto"<<T[u][i].v<<"ww"<<i<<endl;
110         //}
111 
112 
113         int v=T[u][i].v,w=T[u][i].w;
114         if(v==fa)continue;
115 
116         int tmplen=mxlen;
117         EDGE tmpremo=remo;
118 
119         if(mxlen<w){
120             mxlen=w;
121             remo=T[u][i];
122         }
123 
124         dfsx(v,u,mxlen,remo,gf);
125 
126         if(!flag){
127             mxlen=tmplen;
128             remo=tmpremo;
129         }else{
130             return;
131         }
132     }
133 }
134 
135 int main(){
136     while(~scanf("%d",&n)){
137 
138         int x;
139         lim=0;
140         tot=0;blkcnt=0;ans=0;du=0;edgcnt=0;root=0;
141 
142         for(int i=0;i<=32;i++){
143             G[i].clear();T[i].clear();
144             hav[i].clear();
145         }
146         du=0;BLK.clear();SSS.clear();DES.clear();
147         memset(belong,0,sizeof belong);idx.clear();
148         flag=0;
149 
150 
151         
152         //idx["Park"]=8;
153         //idx["1"]=1;idx["2"]=2;
154         //idx["3"]=3;idx["4"]=4;
155         //idx["5"]=5;idx["6"]=6;
156         //idx["7"]=7;idx["8"]=8;
157         //tot=8;    
158 
159 
160 
161         for(int i=1;i<=n;i++){
162             scanf("%s%s%d",buf[0]+1,buf[1]+1,&x);
163 
164             
165             if(idx[buf[0]+1]==0){
166                 idx[buf[0]+1]=++tot;
167             }
168             if(idx[buf[1]+1]==0){
169                 idx[buf[1]+1]=++tot;
170             }
171             
172 
173             int u=idx[buf[0]+1],v=idx[buf[1]+1];
174 
175             //cout<<u<<"ww"<<v<<endl;
176 
177             int w=x;
178 
179             G[idx[buf[0]+1]].pb((EDGE){u,v,w,false,edgcnt++});
180             G[idx[buf[1]+1]].pb((EDGE){v,u,w,false,edgcnt++});
181         }
182         scanf("%d",&lim);
183 
184         root=idx["Park"];
185         for(int i=1;i<=tot;i++){
186             if(!belong[i] && i!=root){
187                 ++blkcnt;
188                 dfs(i,i);
189             }
190         }
191         
192         init(tot);
193 
194         for(int i=1;i<=blkcnt;i++){
195             int mnlen=inf,toid=0;
196 
197             BLK.clear();
198             for(int j=0;j<hav[i].size();j++){
199                 for(int k=0;k<G[hav[i][j]].size();k++){
200                     int v=G[hav[i][j]][k].v,w=G[hav[i][j]][k].w;
201                     if(v!=root)BLK.pb(G[hav[i][j]][k]);
202                     else{
203                         if(w<mnlen){
204                             mnlen=w;
205                             toid=hav[i][j];
206                         }
207                     }
208                 }
209             }
210 
211             sort(BLK.begin(),BLK.end());
212 
213             for(int j=0;j<BLK.size();j++){
214                 int x=BLK[j].u,y=BLK[j].v;
215                 x=find(x);y=find(y);
216                 if(x==y)continue;
217                 else{
218                     unite(x,y);
219                     ans+=BLK[j].w;
220                     
221                     x=BLK[j].u;
222                     y=BLK[j].v;
223 
224                     T[x].pb((EDGE){x,y,BLK[j].w,false,edgcnt++});
225                     T[y].pb((EDGE){y,x,BLK[j].w,false,edgcnt++});
226 
227                     //cout<<x<<" "<<y<<" "<<BLK[j].w<<endl;
228                 }
229             }
230 
231             //cout<<toid<<"www"<<mnlen<<endl;
232 
233             //cout<<ans<<"ww"<<endl;exit(0);
234             //选代表元素
235             T[root].pb((EDGE){root,toid,mnlen,false,edgcnt++});
236             T[toid].pb((EDGE){toid,root,mnlen,false,edgcnt++});
237 
238             ans+=mnlen;
239             SSS.insert(toid);
240             du++;
241         }
242 
243         while(du<lim){
244             priority_queue<NODE>Q;
245 
246             for(int i=0;i<G[root].size();i++){        
247                 
248                 int v=G[root][i].v,w=G[root][i].w;
249                 if(SSS.count(v))continue;
250                 EDGE plc;
251                 int vvl=0;
252 
253                 flag=0;
254 
255                 dfsx(v,v,vvl,plc,v);    //v到root的路径找最长边
256                 
257                 //if(ans==82){
258                 //    cout<<v<<"ww"<<plc.u<<"__"<<plc.v<<endl;
259                 //}
260 
261                 Q.push((NODE){plc,vvl-w,v,w});
262             }
263             if(Q.empty())break;
264             NODE now=Q.top();
265             if(now.val<0)break;
266             ans-=now.val;
267 
268             //cout<<now.to<<"ww"<<now.newLen<<"ww"<<now.val<<"ww"<<ans<<endl;
269             //cout<<now.del.u<<"ww"<<now.del.v<<endl;
270 
271             T[root].pb((EDGE){root,now.to,now.newLen,edgcnt++});
272             T[now.to].pb((EDGE){now.to,root,now.newLen,edgcnt++});
273 
274             //cout<<du<<"du"<<now.to<<"qqq"<<root<<endl;
275 
276             SSS.insert(now.to);
277             DES.insert(now.del.id);
278             DES.insert(now.del.id^1);
279             
280             du++;
281         }
282         printf("Total miles driven: %d\n",ans);
283     }    
284 }
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posted @ 2018-03-27 11:10  Drenight  阅读(269)  评论(0)  编辑  收藏  举报