Sichuan State Programming Contest 2012 Journey
2014-11-18 22:49
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题意:给一棵树,增加一条边后问点对(u,v)的距离减少了多少?
解法:在线LCA,可以用RMQ,或者SPFA(感觉这个有点屌)
具体对一个询问(U,V),未加边时的距离直接就是dis【u】+dis【v】-dis【LCA(u,v)】;
在加边之后,无非就是必须经过这条边的时候u与v距离是多少,即u->x->y>v和u->y->x->v中的最小值。
不知道为毛离线就会超内存,晕~~~。
代码:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
const int NN=100005;
int n,rt;
vector<pair<int,int> > edge[NN];
int depth;
int bn,b[NN*2];
int f[NN*2];
int p[NN];
int dis[NN];
void dfs(int u,int fa)
{
int tmp=++depth;
b[++bn]=tmp; f[tmp]=u; p[u]=bn;
for (int i=0; i<edge[u].size(); i++)
{
int v=edge[u][i].first;
if (v==fa) continue;
dis[v]=dis[u]+edge[u][i].second;
dfs(v,u);
b[++bn]=tmp;
}
}
int dp[NN*2][20];
void rmq_init(int n)
{
for (int i=1; i<=n; i++) dp[i][0]=b[i];
int m=floor(log(n*1.0)/log(2.0));
for (int j=1; j<=m; j++)
for (int i=1; i<=n-(1<<j)+1; i++)
dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int rmq(int l,int r)
{
int k=floor(log((r-l+1)*1.0)/log(2.0));
return min(dp[l][k],dp[r-(1<<k)+1][k]);
}
int lca(int a,int b)
{
if (p[a]>p[b]) swap(a,b);
int k=rmq(p[a],p[b]);
return f[k];
}
inline int get_dis(int i,int j){
return dis[i]+dis[j]-2*dis[lca(i,j)];
}
int main()
{
int q,u,v,w,u1,v1,w1;
int T,ca=0;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&q);
for (int i=1; i<=n; i++) edge[i].clear();
bn=depth=0;
for(int i=0;i<n-1;i++)
{
scanf("%d%d%d",&u,&v,&w);
edge[u].push_back(make_pair(v,w));
edge[v].push_back(make_pair(u,w));
}
scanf("%d%d%d",&u1,&v1,&w1);
rt=1; dis[rt]=0;
dfs(1,0);
rmq_init(bn);
printf("Case #%d:\n",++ca);
while (q--)
{
scanf("%d%d",&u,&v);
int _min=w1+min(get_dis(u,u1)+get_dis(v,v1),get_dis(u,v1)+get_dis(v,u1));
int now=get_dis(u,v);
if(_min>=now) puts("0");
else printf("%d\n",now-_min);
}
}
return 0;
}
解法:在线LCA,可以用RMQ,或者SPFA(感觉这个有点屌)
具体对一个询问(U,V),未加边时的距离直接就是dis【u】+dis【v】-dis【LCA(u,v)】;
在加边之后,无非就是必须经过这条边的时候u与v距离是多少,即u->x->y>v和u->y->x->v中的最小值。
不知道为毛离线就会超内存,晕~~~。
代码:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <cmath>
#include <vector>
using namespace std;
const int NN=100005;
int n,rt;
vector<pair<int,int> > edge[NN];
int depth;
int bn,b[NN*2];
int f[NN*2];
int p[NN];
int dis[NN];
void dfs(int u,int fa)
{
int tmp=++depth;
b[++bn]=tmp; f[tmp]=u; p[u]=bn;
for (int i=0; i<edge[u].size(); i++)
{
int v=edge[u][i].first;
if (v==fa) continue;
dis[v]=dis[u]+edge[u][i].second;
dfs(v,u);
b[++bn]=tmp;
}
}
int dp[NN*2][20];
void rmq_init(int n)
{
for (int i=1; i<=n; i++) dp[i][0]=b[i];
int m=floor(log(n*1.0)/log(2.0));
for (int j=1; j<=m; j++)
for (int i=1; i<=n-(1<<j)+1; i++)
dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
}
int rmq(int l,int r)
{
int k=floor(log((r-l+1)*1.0)/log(2.0));
return min(dp[l][k],dp[r-(1<<k)+1][k]);
}
int lca(int a,int b)
{
if (p[a]>p[b]) swap(a,b);
int k=rmq(p[a],p[b]);
return f[k];
}
inline int get_dis(int i,int j){
return dis[i]+dis[j]-2*dis[lca(i,j)];
}
int main()
{
int q,u,v,w,u1,v1,w1;
int T,ca=0;
scanf("%d",&T);
while(T--){
scanf("%d%d",&n,&q);
for (int i=1; i<=n; i++) edge[i].clear();
bn=depth=0;
for(int i=0;i<n-1;i++)
{
scanf("%d%d%d",&u,&v,&w);
edge[u].push_back(make_pair(v,w));
edge[v].push_back(make_pair(u,w));
}
scanf("%d%d%d",&u1,&v1,&w1);
rt=1; dis[rt]=0;
dfs(1,0);
rmq_init(bn);
printf("Case #%d:\n",++ca);
while (q--)
{
scanf("%d%d",&u,&v);
int _min=w1+min(get_dis(u,u1)+get_dis(v,v1),get_dis(u,v1)+get_dis(v,u1));
int now=get_dis(u,v);
if(_min>=now) puts("0");
else printf("%d\n",now-_min);
}
}
return 0;
}
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