【TYVJ 五月图论专项有奖比赛】

最短路+TSP+最小生成树+倍增LCA+TreeDP

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第一题

　　其实是个TSP问题（然而我没发现），但是关键点很少，只有5个，所以用dij+heap分别预处理出来这五个点为源的最短路……

　　然后枚举起点 i ，枚举这5个点的经过顺序，然后O(1)处理答案就可以了……

　　容易写错的地方是 五个点的标号(a[i])，以及第几个点(i)，这个地方容易搞混……

　　我爆零的地方是：有个地方原来写的是 i 循环，然而我修改了以后内层变成 j 循环了，但是我没改循环内的语句……

　　还有就是数组应该开 len[M<<1]的，我开成len[N<<1] 了。

 #include<queue>
 #include<vector>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 using namespace std;

 int getint(){
     int v=,sign=; char ch=getchar();
     while(ch<''||ch>'') {if (ch=='-') sign=-; ch=getchar();}
     while(ch>=''&&ch<='') {v=v*+ch-''; ch=getchar();}
     return v*sign;
 }
 typedef long long LL;
 const int N=,M=,INF=~0u>>;
 /*******************template********************/
 int to[M<<],next[M<<],len[M<<],head[N],cnt;
 void add(int x,int y,int z){
     to[++cnt]=y; next[cnt]=head[x]; head[x]=cnt; len[cnt]=z;
     to[++cnt]=x; next[cnt]=head[y]; head[y]=cnt; len[cnt]=z;
 }

 int n,m,k;
 int a[],b[],d[][N];
 bool vis[N],sign[N];
 struct node{int x,d;};
 bool operator < (node a,node b){return a.d>b.d;}
 priority_queue<node>Q;

 void dij(int s){
     memset(d[s],0x3f,sizeof d[s]);
     memset(vis,,sizeof vis);
     d[s][a[s]]=;
     Q.push((node){a[s],});
     while(!Q.empty()){
         int x=Q.top().x; Q.pop();
         if (vis[x]) continue;
         vis[x]=;
         for(int i=head[x];i;i=next[i])
             if (d[s][to[i]]>d[s][x]+len[i]){
                 d[s][to[i]]=d[s][x]+len[i];
                 Q.push((node){to[i],d[s][to[i]]});
             }
     }
 }
 int ans=1e9+;
 void check(){
     int tmp=;
     F(i,,n){
         tmp=;
         if (!sign[i]){
             tmp+=d[b[]][i]+d[b[k]][i];
             F(j,,k-) tmp+=d[b[j]][ a[b[j+]] ];
             ans=min(ans,tmp);
         }
     }
 }
 bool chk[];
 void dfs(int x){
     if (x==k+) check();
     F(i,,k) if (!chk[i]){
         b[x]=i;
         chk[i]=;
         dfs(x+);
         b[x]=;
         chk[i]=;
     }
 }
 int main(){
     n=getint(); m=getint(); k=getint();
     F(i,,k){
         a[i]=getint();
         sign[a[i]]=;
     }
     F(i,,m){
         int x=getint(),y=getint(),z=getint();
         add(x,y,z);
     }
     F(i,,k) dij(i);
     dfs();
     printf("%d\n",ans);
     return ;
 }

第二题

　　跟NOIP 2012的d1t3（希望没记错时间……那题叫货车运输……）几乎完全相同的题，只不过那题是找最小，这题是最大。

　　搞一个最大生成树，然后在树上倍增就可以了。

　　40分？……因为忘记判impossible了……2333

 #include<vector>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 using namespace std;
 typedef long long LL;
 inline int getint(){
     int r=,v=; char ch=getchar();
     for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-;
     for(; isdigit(ch);ch=getchar()) v=v*-''+ch;
     return r*v;
 }
 const int N=1e5+,M=3e5+;
 /*******************template********************/

 int to[N<<],next[N<<],head[N],len[N<<],cnt;
 void add(int x,int y,int z){
     to[++cnt]=y; next[cnt]=head[x]; head[x]=cnt; len[cnt]=z;
     to[++cnt]=x; next[cnt]=head[y]; head[y]=cnt; len[cnt]=z;
 }
 struct edge{int u,v,w;}E[M];
 bool cmp(const edge &a,const edge &b){return a.w<b.w;}
 int f[N],sz[N];
 inline int getf(int x){return f[x]==x ? x : getf(f[x]);}

 int fa[N][],mx[N][],dep[N],n,m;
 void dfs(int x){
     F(i,,)
         if (dep[x]>=(<<i)){
             fa[x][i]=fa[fa[x][i-]][i-];
             mx[x][i]=max(mx[fa[x][i-]][i-],mx[x][i-]);
         }else break;
     for(int i=head[x];i;i=next[i])
         if (to[i]!=fa[x][]){
             fa[to[i]][]=x;
             dep[to[i]]=dep[x]+;
             mx[to[i]][]=len[i];
             dfs(to[i]);
         }
 }
 int query(int x,int y){
     if (dep[x]<dep[y]) swap(x,y);
     int t=dep[x]-dep[y];
     int ans=;
     F(i,,) if (t&(<<i)){
         ans=max(mx[x][i],ans);
         x=fa[x][i];
     }
     D(i,,) if (fa[x][i]!=fa[y][i]){
         ans=max(mx[x][i],ans);
         ans=max(mx[y][i],ans);
         x=fa[x][i]; y=fa[y][i];
     }
     if (x!=y) ans=max(ans,max(mx[x][],mx[y][]));
     return ans;
 }
 int main(){
     n=getint(); m=getint();
     F(i,,m){
         int x=getint(),y=getint(),z=getint();
         E[i]=(edge){x,y,z};
     }
     sort(E+,E+m+,cmp);
     F(i,,n) f[i]=i,sz[i]=;
     F(i,,m){
         int f1=getf(E[i].u),f2=getf(E[i].v);
         if (f1!=f2){
             if (sz[f1]>sz[f2]) swap(f1,f2);
             f[f1]=f2;
             sz[f2]+=sz[f1];
             add(E[i].u,E[i].v,E[i].w);
         }
     }

     dfs();
     int T=getint();
     while(T--){
         int x=getint(),y=getint();
         if (getf(x)!=getf(y)) puts("impossible");
         else printf("%d\n",query(x,y));
     }
     return ;
 }

第三题

　　我并不会写……只好预处理出来dist[i][j]数组，然后$N^2$枚举+O(n)判断……三次方的做法骗了30分= =

　　然而正解是找树的重心？预处理出来每棵子树的答案，由于整棵树分成的两部分是一棵完整的子树+剩下的部分，所以枚举那个砍掉的子树？在再加上一些神奇的技巧……

　　然而一上午也没yy出来aaarticlea/gif;base64,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" alt="" />果然代码能力就是渣渣，sad……

　　UPD：2015年5月20日 16:28:33

　　Orz zyf神犇

　　果然是……枚举断了哪条边，然后暴力修改= =用dfs整棵树来选边就可以，从当前点沿fa修改上去，两半树各求一遍答案……ok……无限ym啊……

 //TYVJ C
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 #define pb push_back
 using namespace std;
 typedef long long LL;
 inline int getint(){
     int r=,v=; char ch=getchar();
     for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-;
     for(; isdigit(ch);ch=getchar()) v=v*-''+ch;
     return r*v;
 }
 const int N=;
 /*******************template********************/
 int to[N<<],next[N<<],head[N],cnt;
 void add(int x,int y){
     to[++cnt]=y; next[cnt]=head[x]; head[x]=cnt;
     to[++cnt]=x; next[cnt]=head[y]; head[y]=cnt;
 }

 int n,a[N],f[N][],dep[N],s[N],fa[N],cut;
 LL g[N],ans=1e15;
 void dfs(int x){
     for(int i=head[x];i;i=next[i])
         if (to[i]!=fa[x]){
             dep[to[i]]=dep[x]+;
             fa[to[i]]=x;
             dfs(to[i]);
             s[x]+=s[to[i]];
             if (s[to[i]]>s[f[x][]]) f[x][]=f[x][],f[x][]=to[i];
             else if (s[to[i]]>s[f[x][]]) f[x][]=to[i];
             g[x]+=g[to[i]]+s[to[i]];
         }
 }
 LL getans(int x,int cnt){
     int t=(f[x][]==cut || s[f[x][]]>s[f[x][]]) ? f[x][] : f[x][];
     if (*s[t]>cnt) return *s[t]-cnt+getans(t,cnt);
     else return ;
 }
 void dp(int x){
     for(int i=head[x];i;i=next[i])
         if (to[i]!=fa[x]){
             cut=to[i];
             for(int j=x;j;j=fa[j]) s[j]-=s[to[i]];
             ans=min(ans,g[]-g[to[i]]-(LL)s[to[i]]*dep[to[i]]-getans(,s[])+g[to[i]]-getans(to[i],s[to[i]]));
             for(int j=x;j;j=fa[j]) s[j]+=s[to[i]];
             dp(to[i]);
         }
 }

 int main(){
 #ifndef ONLINE_JUDGE
     freopen("C.in","r",stdin);
     freopen("C.out","w",stdout);
 #endif
     n=getint();
     F(i,,n){
         int x=getint(),y=getint();
         add(x,y);
     }
     F(i,,n) s[i]=getint();
     dfs();
     dp();
     printf("%lld\n",ans);
     return ;
 }