水题赛无题解
sol：可以用权值并查集搞，但感觉还是建图跑bfs比较容易
定义S[i]表示前缀和
读入a，b，c；就是S[a-1] + c=S[b]，那就从a-1向b连一条c，b向a-1连一条-c
但这样有可能这张图并不联通，于是每次bfs时给节点染色，颜色不同显然to hard
询问随便处理下就可以了
/* S[3]-S[0]=10 S[3]-S[2]=5 */ #include <bits/stdc++.h> using namespace std; #define int long long typedef int ll; inline ll read() { ll S=0; bool f=0; char ch=' '; while(!isdigit(ch)) { f|=(ch=='-'); ch=getchar(); } while(isdigit(ch)) { S=(S<<3)+(S<<1)+(ch-'0'); ch=getchar(); } return (f)?(-S):(S); } #define R(x) x=read() inline void write(ll x) { if(x<0) { putchar('-'); x=-x; } if(x<10) { putchar(x+'0'); return; } write(x/10); putchar((x%10)+'0'); return; } #define W(x) write(x),putchar(' ') #define Wl(x) write(x),putchar(' ') const int N=100005,M=200005; int n,m,Q,Cnt=0; namespace Dijkstra { int tot=0,Next[M],to[M],val[M],head[N]; int Cor[N]; inline void add(int x,int y,int z) { Next[++tot]=head[x]; to[tot]=y; val[tot]=z; head[x]=tot; return; } struct Record { int Weiz,Val; }; inline bool operator<(Record p,Record q) { return p.Val>q.Val; } priority_queue<Record>Queue; bool Arr[N]; int Dis[N]; inline void Run(int S,int Co) { Cor[S]=Co; Queue.push((Record){S,Dis[S]=0}); int i; while(!Queue.empty()) { Record tmp=Queue.top(); Cor[tmp.Weiz]=Co; Queue.pop(); if(Arr[tmp.Weiz]) continue; Arr[tmp.Weiz]=1; for(i=head[tmp.Weiz];i;i=Next[i]) { if(Dis[to[i]]>tmp.Val+val[i]) { Dis[to[i]]=tmp.Val+val[i]; if(!Arr[to[i]]) Queue.push((Record){to[i],Dis[to[i]]}); } } } } inline void Solve(int x,int y) { y++; if(Cor[x]!=Cor[y]) { puts("Too Hard"); } else { Wl(Dis[y]-Dis[x]); } } inline void Init() { memset(head,0,sizeof head); memset(Arr,0,sizeof Arr); memset(Dis,63,sizeof Dis); memset(Cor,0,sizeof Cor); return; } } signed main() { freopen("sing.in","r",stdin); freopen("sing.out","w",stdout); R(n); R(m); int i; Dijkstra::Init(); for(i=1;i<=m;i++) { int x=read(),y=read()+1,z=read(); Dijkstra::add(x,y,z); Dijkstra::add(y,x,-z); } for(i=1;i<=n;i++) if(!Dijkstra::Arr[i]) { Dijkstra::Run(i,++Cnt); } R(Q); for(i=1;i<=Q;i++) { int x=read(),y=read(); Dijkstra::Solve(x,y); } return 0; } /* input 4 2 1 3 10 3 3 5 2 1 2 1 4 output 5 Too Hard */
sol：容易发现这个n非常大，肯定不能直接上最短路
于是发现跑最短路时，真正有用的是那些中间有墙倒掉的点，对于这2*m个点
首先在有墙倒的地方建一条权值1的双向边
在对2*m个点排序，p[i]向p[i+1]连一条权值为(p[i+1]-p[i])的边
要离散。。。
#include <bits/stdc++.h> using namespace std; #define int long long typedef int ll; inline ll read() { ll S=0; bool f=0; char ch=' '; while(!isdigit(ch)) { f|=(ch=='-'); ch=getchar(); } while(isdigit(ch)) { S=(S<<3)+(S<<1)+(ch-'0'); ch=getchar(); } return (f)?(-S):(S); } #define R(x) x=read() inline void write(ll x) { if(x<0) { putchar('-'); x=-x; } if(x<10) { putchar(x+'0'); return; } write(x/10); putchar((x%10)+'0'); return; } #define W(x) write(x),putchar(' ') #define Wl(x) write(x),putchar(' ') const int N=500005,M=1000005; int n,m; int X[N],Y[N]; int P[N<<1]; int Pos[N<<1]; map<int,int>Map; namespace Dijkstra { int tot=0,Next[M],to[M],val[M],head[N]; inline void add(int x,int y,int z) { Next[++tot]=head[x]; to[tot]=y; val[tot]=z; head[x]=tot; return; } struct Record { int Weiz,Val; }; inline bool operator<(Record p,Record q) { return p.Val>q.Val; } priority_queue<Record>Queue; bool Arr[N]; int Dis[N]; inline void Run(int S) { Queue.push((Record){S,Dis[S]=0}); int i; while(!Queue.empty()) { Record tmp=Queue.top(); Queue.pop(); if(Arr[tmp.Weiz]) continue; Arr[tmp.Weiz]=1; for(i=head[tmp.Weiz];i;i=Next[i]) { if(Dis[to[i]]>tmp.Val+val[i]) { Dis[to[i]]=tmp.Val+val[i]; if(!Arr[to[i]]) Queue.push((Record){to[i],Dis[to[i]]}); } } } } inline void Init() { memset(head,0,sizeof head); memset(Arr,0,sizeof Arr); memset(Dis,63,sizeof Dis); return; } } #define Dj Dijkstra signed main() { freopen("lh.in","r",stdin); freopen("lh.out","w",stdout); R(n); R(m); int i; Dj::Init(); for(i=1;i<=m;i++) { X[i]=read(); Y[i]=read(); if(!Map[X[i]]) { P[Map[X[i]]=++*P]=X[i]; } if(!Map[Y[i]]) { P[Map[Y[i]]=++*P]=Y[i]; } Dj::add(Map[X[i]],Map[Y[i]],1); Dj::add(Map[Y[i]],Map[X[i]],1); } if(!Map[1]) { P[Map[1]=++*P]=1; } if(!Map[n]) { P[Map[n]=++*P]=n; } sort(P+1,P+*P+1); for(i=1;i<*P;i++) { Dj::add(Map[P[i]],Map[P[i+1]],P[i+1]-P[i]); Dj::add(Map[P[i+1]],Map[P[i]],P[i+1]-P[i]); } Dj::Run(Map[1]); Wl(Dj::Dis[Map[n]]); return 0; } /* input 31 9 4 15 10 25 13 30 1 6 2 9 6 19 9 24 10 27 1 4 output 6 Too Hard */
sol：看上去很厉害其实只是MST的板子罢了，一堆描述只是在故弄玄虚，就是说没有环。
Prim随便水
#include <bits/stdc++.h> using namespace std; typedef int ll; inline ll read() { ll S=0; bool f=0; char ch=' '; while(!isdigit(ch)) { f|=(ch=='-'); ch=getchar(); } while(isdigit(ch)) { S=(S<<3)+(S<<1)+(ch-'0'); ch=getchar(); } return (f)?(-S):(S); } #define R(x) x=read() inline void write(ll x) { if(x<0) { putchar('-'); x=-x; } if(x<10) { putchar(x+'0'); return; } write(x/10); putchar((x%10)+'0'); return; } #define W(x) write(x),putchar(' ') #define Wl(x) write(x),putchar(' ') const int N=5005; int n; struct Point { int x,y; }P[N]; double DD[N]; int Father[N]; bool Vis[N]; inline double Calc(int i,int j) { double ix=P[i].x,iy=P[i].y,jx=P[j].x,jy=P[j].y; return sqrt((ix-jx)*(ix-jx)+(iy-jy)*(iy-jy)); } int main() { freopen("torture.in","r",stdin); freopen("torture.out","w",stdout); int i,j; R(n); for(i=1;i<=n;i++) { Father[i]=i; R(P[i].x); R(P[i].y); } int step; double ans=0; for(i=2;i<=n;i++) DD[i]=1e15; DD[1]=0; for(step=1;step<n;step++) { int u=-1; for(i=1;i<=n;i++) if((!Vis[i])&&(u==-1||DD[i]<DD[u])) u=i; Vis[u]=1; for(i=1;i<=n;i++) if((!Vis[i])&&(DD[i]>Calc(u,i))) { DD[i]=Calc(u,i); } } for(i=1;i<=n;i++) ans+=DD[i]; printf("%.2lf ",ans); return 0; } /* input 4 0 0 1 2 -1 2 0 4 output 6.47 */
updata：首先对于题面有一处修正，两个点之间的距离是min（|Xa-Xb|,|Ya-Yb|,|Za-Zb|）,题面中少了个min，而且也不是曼哈顿距离
sol：对三维分别排序，得到3*（n-1）条边，Kruskal直接上
Xjb证明：只有一维时，显然排好序后第i个向i+1连边最优，
当有三维时，Kruskal保证当前边权是没选过的中最小的，相当于已经在3维中去过min了，于是可以直接做了
#include <bits/stdc++.h> using namespace std; typedef long long ll; inline ll read() { ll S=0; bool f=0; char ch=' '; while(!isdigit(ch)) { f|=(ch=='-'); ch=getchar(); } while(isdigit(ch)) { S=(S<<3)+(S<<1)+(ch-'0'); ch=getchar(); } return (f)?(-S):(S); } #define R(x) x=read() inline void write(ll x) { if(x<0) { putchar('-'); x=-x; } if(x<10) { putchar(x+'0'); return; } write(x/10); putchar((x%10)+'0'); return; } #define W(x) write(x),putchar(' ') #define Wl(x) write(x),putchar(' ') const int N=100005; int n; int Father[N]; struct Point { int Weiz,Id; }X[N],Y[N],Z[N]; inline bool Point_cmp(Point a,Point b) { return a.Weiz<b.Weiz; } int cnt=0; struct Edge { int U,V,Bianquan; }E[N*3]; inline bool Edge_cmp(Edge a,Edge b) { return a.Bianquan<b.Bianquan; } inline int Get_Father(int x) { return (Father[x]==x)?(x):(Father[x]=Get_Father(Father[x])); } int main() { freopen("time.in","r",stdin); freopen("time.out","w",stdout); int i; R(n); for(i=1;i<=n;i++) { X[i]=(Point){read(),i}; Y[i]=(Point){read(),i}; Z[i]=(Point){read(),i}; } sort(X+1,X+n+1,Point_cmp); for(i=1;i<n;i++) { E[++cnt]=(Edge){X[i].Id,X[i+1].Id,X[i+1].Weiz-X[i].Weiz}; } sort(Y+1,Y+n+1,Point_cmp); for(i=1;i<n;i++) { E[++cnt]=(Edge){Y[i].Id,Y[i+1].Id,Y[i+1].Weiz-Y[i].Weiz}; } sort(Z+1,Z+n+1,Point_cmp); for(i=1;i<n;i++) { E[++cnt]=(Edge){Z[i].Id,Z[i+1].Id,Z[i+1].Weiz-Z[i].Weiz}; } sort(E+1,E+cnt+1,Edge_cmp); for(i=1;i<=n;i++) { Father[i]=i; } ll ans=0; int Bians=0; for(i=1;i<=cnt;i++) { int u=E[i].U,v=E[i].V; int uu=Get_Father(u),vv=Get_Father(v); if(uu==vv) continue; ans+=1ll*E[i].Bianquan; Father[uu]=vv; if(++Bians==n-1) break; } Wl(ans); return 0; } /* input 5 11 -15 -15 14 -5 -15 -1 -1 -5 10 -4 -1 19 -4 19 output 4 */