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【CF1051F】The Shortest Statement 题解

Educational Codeforces Round 51 (Rated for Div. 2)E题

也许是一道码题

核心思路

直接floyd是肯定过不了的,注意到mn20m-n\leq20,那么这个图可以看做为一棵树和至多20条非树边构成的。那么在树上求距离就很简单了,对于剩下的20条非树边直接暴力从两个端点开始求一遍最短路,在所有经过至少一条非树边与树上距离中找到最小值即为答案。

uvu\to v至少经过一条(u0,v0)(u_0,v_0)的边的最小值为min{distid(u0)u+distid(v0)v}\min \{\mathrm{dist}_{id(u_0)}u+\mathrm{dist}_{id(v_0)}v\}

完整代码

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#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<queue>
using namespace std;
#define int ll

#define REP(i,e,s) for(register int i=e; i<=s; i++)
#define DREP(i,e,s) for(register int i=e; i>=s; i--)
#define ll long long
#define DE(...) fprintf(stderr,__VA_ARGS__)
#define DEBUG(a) DE("DEBUG: %d\n",a)
#define file(a) freopen(a".in","r",stdin);freopen(a".out","w",stdout)
int read() {
int x=0,f=1,ch=getchar();
while(ch>'9'||ch<'0'){if(ch=='-')f=-1;ch=getchar();}
while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
return x*f;
}

const int MAXN=200000+10,INF=0x3f3f3f3f3f3f3f;

struct edge {
int u,v,w;
} a[MAXN];

int f[MAXN];

int find(int x) {
if(f[x]==x) return x;
return f[x]=find(f[x]);
}

int link(int x,int y) {
f[find(x)]=find(y);
}

int head[MAXN],_next[MAXN<<1],to[MAXN<<1],weigh[MAXN<<1],cnt;

void addedge(int u,int v,int w) {
cnt++;
_next[cnt]=head[u];
head[u]=cnt;
to[cnt]=v;
weigh[cnt]=w;
}

queue<int> q1;

int n,m,dept[MAXN],dis[MAXN],fa[MAXN][25];

void bfs(int s) {
q1.push(s);
while(!q1.empty()) {
int u=q1.front();q1.pop();
for(int i=head[u]; i; i=_next[i]) {
int v=to[i];
if(fa[u][0]==v) continue;
fa[v][0]=u;
dept[v]=dept[u]+1;
dis[v]=dis[u]+weigh[i];
q1.push(v);
}
}
}

priority_queue<pair<int,int> > q;

int book[MAXN],dist[44][MAXN];

void dij(int op,int s) {
q.push(make_pair(0,s));
REP(i,1,n) book[i]=0;
REP(i,1,n) dist[op][i]=INF;
dist[op][s]=0;
while(!q.empty()) {
int u=q.top().second;q.pop();
if(book[u]) continue;
book[u]=1;
for(int i=head[u]; i; i=_next[i]) {
int v=to[i];
if(dist[op][v]>dist[op][u]+weigh[i]) {
dist[op][v]=dist[op][u]+weigh[i];
q.push(make_pair(-dist[op][v],v));
}
}
}
return ;
}

int used[MAXN];

int lca(int u,int v) {
if(dept[u]>dept[v]) swap(u,v);
int len=dept[v]-dept[u];
DREP(i,19,0) if((1<<i)&len) v=fa[v][i];
if(u==v) return u;
DREP(i,19,0) if(fa[u][i]!=fa[v][i]) u=fa[u][i],v=fa[v][i];
return fa[u][0];
}

signed main() {
n=read(),m=read();
REP(i,1,m)
a[i].u=read(),a[i].v=read(),a[i].w=read();

REP(i,1,n) f[i]=i;

int num=0;

REP(i,1,m) {
if(find(a[i].u)==find(a[i].v)) continue;
link(a[i].u,a[i].v);
addedge(a[i].u,a[i].v,a[i].w);
addedge(a[i].v,a[i].u,a[i].w);
used[i]=1;
num++;
if(num==n-1) break;
}
bfs(1);
REP(j,1,19) REP(i,1,n) fa[i][j]=fa[fa[i][j-1]][j-1];


REP(i,1,m)
if(!used[i])
addedge(a[i].u,a[i].v,a[i].w),addedge(a[i].v,a[i].u,a[i].w);
int op=0;
REP(i,1,m) {
if(!used[i]) {
dij(++op,a[i].u);
dij(++op,a[i].v);
}
}
int q=read();
REP(i,1,q) {
int u=read(),v=read();
int ans=dis[u]+dis[v]-2*dis[lca(u,v)];
REP(j,1,op) ans=min(ans,dist[j][u]+dist[j][v]);
printf("%lld\n",ans);
}

return 0;
}