【Codeforces 609E】【Stree】【最小生成树】

题目大意

给出n个点,m条有权边,现对于每一条边，你需要回答出包含这条边的最小生成树的总边权值。

题解

一个显然的结论，无论怎么样，生成树与最小生成树不同的边最多为一条。求出最小生成树，枚举加入哪条边，用倍增算法求出所加边在树上路径边权的最大值，更改答案即可。

code

#include<set>#include<cmath>#include<cstdio>#include<cstring>#include<algorithm>#define fo(i,j,k) for(int i=j;i<=k;i++)#define fd(i,j,k) for(int i=j;i>=k;i--)using namespace std;int const maxn=200000,maxm=200000;int n,m,gra,to[maxn*2+10],val[maxn*2+10],next[maxn*2+10],begin[maxn+10],dep[maxn+10],    father[maxn+10][20],mx[maxn+10][20],fa[maxn+10],queue[maxn+10];bool inqueue[maxn+10];long long ans[maxm+10];struct rec{int u,v,w,num;};rec edge[maxm+10];bool cmp(rec i,rec j){    return (i.w<j.w)||((i.w==j.w)&&(i.u<j.u))||((i.w==j.w)&&(i.u==j.u)&&(i.v<j.v));}int getfather(int x){    if(!fa[x])return x;    return fa[x]=getfather(fa[x]);}void insert(int u,int v,int w){    to[++gra]=v;    val[gra]=w;    next[gra]=begin[u];    begin[u]=gra;}void bfs(int s){    int head=0,tail=0;    inqueue[queue[++tail]=s]=1;    for(;head!=tail;){        int now=queue[++head];        for(int i=begin[now];i;i=next[i])            if(!inqueue[to[i]]){                dep[to[i]]=dep[now]+1;                father[to[i]][0]=now;                mx[to[i]][0]=val[i];                inqueue[queue[++tail]=to[i]]=1;            }    }}int lc(int u,int v){    int ans=0;    if(dep[u]<dep[v])swap(u,v);    fd(i,log(n)/log(2),0)        if(dep[father[u][i]]>=dep[v]){            ans=max(ans,mx[u][i]);            u=father[u][i];        }    if(u==v)return ans;    fd(i,log(n)/log(2),0)        if(father[u][i]!=father[v][i]){            ans=max(ans,max(mx[u][i],mx[v][i]));            u=father[u][i];            v=father[v][i];        }    return max(ans,max(mx[u][0],mx[v][0]));}int main(){    //freopen("street.in","r",stdin);    //freopen("street.out","w",stdout);    freopen("d.in","r",stdin);    freopen("d.out","w",stdout);    scanf("%d%d",&n,&m);    fo(i,1,m)        scanf("%d%d%d",&edge[i].u,&edge[i].v,&edge[i].w),edge[i].num=i;    sort(edge+1,edge+m+1,cmp);    fo(i,1,m){        int fu=getfather(edge[i].u),fv=getfather(edge[i].v);        if(fu!=fv){            ans[0]+=edge[i].w;            ans[edge[i].num]=-1;            fa[fu]=fv;            insert(edge[i].u,edge[i].v,edge[i].w);            insert(edge[i].v,edge[i].u,edge[i].w);        }    }    dep[1]=1;    bfs(1);    fo(j,1,log(n)/log(2))        fo(i,1,n){            father[i][j]=father[father[i][j-1]][j-1];            mx[i][j]=max(mx[i][j-1],mx[father[i][j-1]][j-1]);        }    fo(i,1,m)        if(ans[edge[i].num]==-1)ans[edge[i].num]=ans[0];        else ans[edge[i].num]=ans[0]+edge[i].w-lc(edge[i].u,edge[i].v);    fo(i,1,m)printf("%lld\n",ans[i]);    return 0;}