https://vijos.org/p/1843
本题 kruskal+倍增lca
WA了三次 T_T
总结出一些注意事项:
1. 一定要想明白每一部分是求最大值还是最小值!!!(WA 1)
2. 注意操作顺序,别把要用的变量的数值改没!!!(WA 2)
3. 仔细看题,如本题让“”货车不能到达目的地,输出-1“”,而货车载重为零则应输出0!!!!!!(WA 3)
#include<stdio.h>
#include<iostream>
#include<string.h>
#include<algorithm>
using namespace std;
struct edg
{
int x;
int y;
int z;
}edge[];
int n,m;
int q;
bool cmp(edg a,edg b)
{
return a.z>b.z;
}
int min(int a,int b){return a<b?a:b;}
int father[];
int find(int a)
{
if(father[a]!=a) father[a]=find(father[a]);
return father[a];
}
void unionn(int a,int b)
{
father[a]=b;
}
int to[],val[],nextt[],head[],cnt;
void add(int a,int b,int c){to[++cnt]=b;val[cnt]=c;nextt[cnt]=head[a];head[a]=cnt;}
int fa[][];
int minn[][];
int deep[];
void add_root()
{
int v[];
memset(v,,sizeof(v));
for(int i=;i<=n;i++)
{
int a=find(i);
if(!v[a])
{
add(n+,a,);
add(a,n+,);
v[a]=;
}
}
}
int visited[];
void dfs(int pos,int pre,int step,int vv)
{
visited[pos]=;
fa[pos][]=pre;deep[pos]=step;
minn[pos][]=vv;
for(int i=head[pos];i;i=nextt[i])
{
if(!visited[to[i]])
{
dfs(to[i],pos,step+,val[i]);
}
}
}
void swap(int &a,int &b){int t=a;a=b;b=t;}
int lca_min(int x,int y)
{
int re;
int ans=;
if(deep[x]<deep[y])swap(x,y);
for(int i=;i>=;i--)
{
if(deep[fa[x][i]]>=deep[y])
{
ans=min(ans,minn[x][i]);
x=fa[x][i];
}
}
if(x==y) return ans;
for(int i=;i>=;i--)
{
if(fa[x][i]!=fa[y][i])
{
ans=min(ans,minn[x][i]);
ans=min(ans,minn[y][i]);
x=fa[x][i];
y=fa[y][i];
}
else
{
re=fa[x][i];
}
}
ans=min(ans,minn[x][]);
ans=min(ans,minn[y][]);
return ans;
}
int main()
{
scanf("%d%d",&n,&m);
for(int i=;i<=m;i++)
{
scanf("%d%d%d",&edge[i].x,&edge[i].y,&edge[i].z);
edge[i].z++;
}
sort(edge+,edge+m+,cmp);
for(int i=;i<=n;i++) father[i]=i;
for(int i=;i<=m;i++)
{
int a=edge[i].x;
int b=edge[i].y;
a=find(a);
b=find(b);
if(a!=b)
{
unionn(a,b);
add(edge[i].x,edge[i].y,edge[i].z);
add(edge[i].y,edge[i].x,edge[i].z);
}
}
memset(minn,,sizeof(minn));
add_root();
dfs(n+,n+,,);
for(int j=;j<=;j++)
{
for(int i=;i<=n;i++)
{
fa[i][j]=fa[fa[i][j-]][j-];
minn[i][j]=min(minn[i][j-],minn[fa[i][j-]][j-]);
}
}
scanf("%d",&q);
for(int i=;i<=q;i++)
{
int a,b;
scanf("%d%d",&a,&b);
int ans=lca_min(a,b);
if(ans==) printf("-1\n");
else printf("%d\n",ans-);
}
return ;
}
附上几组数据:
/*
5 8
5 4 1
4 2 3
5 3 4
3 1 6
5 1 6
4 2 0
4 2 0
1 2 9
5
1 5
4 3
1 4
4 1
3 4
*/
/*
5 5
2 5 3
5 3 0
1 2 9
3 2 0
4 2 8
5
3 2
2 1
3 4
5 2
1 2
*/
/*
6 8
5 3 7
3 2 3
6 5 2
2 3 6
1 3 6
4 1 3
2 6 9
4 1 1
1
6 4
*/