版权声明:本文为博主原创文章,遵循 CC 4.0 BY-SA 版权协议,转载请附上原文出处链接和本声明。
安全出行Safe Travel
题意:
这道题最开始很容易想成最短路再减去一条边
但是交了后只有10 分 才意识到问题没有那摩简单
其实这道题是最短路生成树 先用dj建一个最短路生成树,再用非树边更新即可
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+100;
const int inf=0x3f3f3f3f;
inline int read()
{
int data=0;int w=1; char ch=0;
ch=getchar();
while(ch!='-' && (ch<'0' || ch>'9')) ch=getchar();
if(ch=='-') w=-1,ch=getchar();
while(ch>='0' && ch<='9') data=(data<<3)+(data<<1)+ch-'0',ch=getchar();
return data*w;
}
struct node{
int u,v,w,nxt;
}e[N<<2];
int fir[N],cnt=0,tfa[N],fa[N];
inline void add(int u,int v,int w){e[++cnt]=(node){u,v,w,fir[u]};fir[u]=cnt;}
struct point{
int u,dist;
bool operator >(const point &t) const{
return dist>t.dist;
}
};
struct t{
int u,v,w;
}tree[N<<2];
int dis[N],n,m,tot=0,ans[N];
priority_queue<point,vector<point>,greater<point> >q;
bool cmp(const t&a,const t&b){
return a.w<b.w;
}
inline int getfa(int x){
return x==fa[x]?x:fa[x]=getfa(fa[x]);
}
inline void dj(int st){
memset(dis,0x3f,sizeof(dis));
dis[st]=0; q.push((point){st,0});
while(!q.empty()){
int u=q.top().u,d=q.top().dist;q.pop();
for(int i=fir[u];i;i=e[i].nxt){
int v=e[i].v;
if(dis[v]>dis[u]+e[i].w){
dis[v]=dis[u]+e[i].w;
tfa[v]=u;
q.push((point){v,dis[v]});
}
}
}
}
int main(){
n=read();m=read();
for(int i=1;i<=m;i++){
int u=read(),v=read(),w=read();
add(u,v,w);add(v,u,w);
}
dj(1);
for(int i=1;i<=2*m;i++){
int u=e[i].u,v=e[i].v;
if(tfa[u]==v||tfa[v]==u) continue;
tree[++tot].w=dis[u]+dis[v]+e[i].w;
tree[tot].u=u;tree[tot].v=v;
}
sort(tree+1,tree+tot+1,cmp);
memset(ans,-1,sizeof(ans));
for(int i=1;i<=n;i++) fa[i]=i;
for(int i=1;i<=tot;i++){
int u=getfa(tree[i].u),v=getfa(tree[i].v);
while(u!=v){
if(dis[u]<dis[v]) swap(u,v);
ans[u]=tree[i].w-dis[u];
fa[u]=tfa[u];
u=getfa(u);
}
}
for(int i=2;i<=n;i++) printf("%d\n",ans[i]);
return 0;
}
还有一个不带路径压缩的
#include<bits/stdc++.h>
using namespace std;
const int N=1e5+100;
const int inf=0x3f3f3f3f;
inline int read()
{
int data=0;int w=1; char ch=0;
ch=getchar();
while(ch!='-' && (ch<'0' || ch>'9')) ch=getchar();
if(ch=='-') w=-1,ch=getchar();
while(ch>='0' && ch<='9') data=(data<<3)+(data<<1)+ch-'0',ch=getchar();
return data*w;
}
struct node{
int u,v,w,nxt;
}e[N<<2];
int fir[N],cnt=0,tfa[N],fa[N];
inline void add(int u,int v,int w){e[++cnt]=(node){u,v,w,fir[u]};fir[u]=cnt;}
struct point{
int u,dist;
bool operator >(const point &t) const{
return dist>t.dist;
}
};
struct t{
int u,v,w;
}tree[N<<2];int ok[N];
int dis[N],n,m,tot=0,ans[N];
priority_queue<point,vector<point>,greater<point> >q;
bool cmp(const t&a,const t&b){
return a.w<b.w;
}
inline int getfa(int x){
return x==fa[x]?x:fa[x]=getfa(fa[x]);
}
inline void dj(int st){
memset(dis,0x3f,sizeof(dis));
dis[st]=0; q.push((point){st,0});
while(!q.empty()){
int u=q.top().u,d=q.top().dist;q.pop();
for(int i=fir[u];i;i=e[i].nxt){
int v=e[i].v;
if(dis[v]>dis[u]+e[i].w){
dis[v]=dis[u]+e[i].w;
tfa[v]=u;
q.push((point){v,dis[v]});
}
}
}
}
int main(){
n=read();m=read();
for(int i=1;i<=m;i++){
int u=read(),v=read(),w=read();
add(u,v,w);add(v,u,w);
}
dj(1);
for(int i=1;i<=2*m;i++){
int u=e[i].u,v=e[i].v;
if(tfa[u]==v||tfa[v]==u) continue;
tree[++tot].w=dis[u]+dis[v]+e[i].w;
tree[tot].u=u;tree[tot].v=v;
}
sort(tree+1,tree+tot+1,cmp);
memset(ans,0x3f,sizeof(ans));
for(int i=1;i<=tot;i++){
int u=tree[i].u,v=tree[i].v;
while(u!=v){
if(dis[u]<dis[v]) swap(u,v);
ans[u]=min(ans[u],tree[i].w-dis[u]);
if(ans[u]!=0)ok[u]=1;
u=tfa[u];
}
}
for(int i=2;i<=n;i++) {
if(ok[i]) printf("%d\n",ans[i]);
else puts("-1");
}
return 0;
}