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(‘\n‘)
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(‘\n‘)
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(‘\n‘)
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\n",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(‘\n‘)
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
*/
原文地址：https://www.cnblogs.com/gaojunonly1/p/10453070.html
时间： 2024-10-03 05:31:01