hdu 6393 Traffic Network in Numazu （树链剖分+线段树 基环树）

链接：http://acm.hdu.edu.cn/showproblem.php?pid=6393

思路：n个点，n条边，也就是基环树。。因为只有一个环，我们可以把这个环断开，建一个新的点n+1与之相连，然后就按照树链剖分求边权的方法分类讨论下，过不过这条被分开的边，一共有三种情况取值最小的。

实现代码：

#include<bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define ll long long
#define mid int m = (l + r) >> 1
const int M = 1e5+10;
int cnt,cnt1,head[M],fa[M],dep[M],son[M],siz[M],top[M],tid[M],n,q,vis[M];
ll val[M];
ll sum[M<<2];
struct node{
    int to,next;
    ll val;
}e[M];

struct node1{
    int u,v;
    ll val;
}a[M];
void add(int u,int v){
    e[++cnt].to = v;e[cnt].next = head[u];head[u] = cnt;
}

void dfs1(int u,int faz,int deep){
    dep[u] = deep;
    fa[u] = faz;
    siz[u] = 1;
    for(int i = head[u];i;i=e[i].next){
        int v = e[i].to;
        if(v == fa[u]) continue;
        dfs1(v,u,deep+1);
        siz[u] += siz[v];
        if(siz[v] > siz[son[u]]||son[u]==-1)
            son[u] = v;
    }
}

void dfs2(int u,int t){
    top[u] = t;
    tid[u] = ++cnt1;
    if(son[u] == -1) return ;
    dfs2(son[u],t);
    for(int i = head[u];i;i=e[i].next){
        int v = e[i].to;
        if(v != fa[u]&&v != son[u])
            dfs2(v,v);

    }
}

void pushup(int rt){
    sum[rt] = sum[rt<<1] + sum[rt<<1|1];
}

void build(int l,int r,int rt){
     if(l == r){
        sum[rt] = val[l];
        return ;
     }
     mid;
     build(lson); build(rson);
     pushup(rt);
}

void update(int p,ll c,int l,int r,int rt){
    if(l == r){
        sum[rt] = c;
        return ;
    }
    mid;
    if(p <= m) update(p,c,lson);
    else update(p,c,rson);
    pushup(rt);
}

ll query(int L,int R,int l,int r,int rt){
    if(L <= l&&R >= r)  return sum[rt];
    mid;
    ll ret = 0;
    if(L <= m) ret += query(L,R,lson);
    if(R > m) ret += query(L,R,rson);
    return ret;
}

/*void ct(int l,int r,int rt){
    if(l == r){
        cout<<sum[rt]<<" ";
        return ;
    }
    mid;
    ct(lson); ct(rson);
}*/

ll solve(int x,int y){
    int fx = top[x],fy = top[y];
    ll ans = 0;
    while(fx != fy){
        if(dep[fx] < dep[fy]) swap(fx,fy),swap(x,y);
        if(fx == 1) ans += query(tid[fx]+1,tid[x],1,n+1,1);
        else ans += query(tid[fx],tid[x],1,n+1,1);
       // cout<<x<<" "<<fx<<" "<<ans<<endl;
        x = fa[fx]; fx = top[x];
    }
    if(x == y) return ans;
    if(dep[x] > dep[y]) swap(x,y);
    ans += query(tid[son[x]],tid[y],1,n+1,1);
    return ans;
}

void init(){
    cnt = cnt1 = 0;
    memset(son,-1,sizeof(son));
    memset(head,0,sizeof(head));
    memset(vis,0,sizeof(vis));
}

int main()
{
    int t;
    scanf("%d",&t);
    while(t--){
        scanf("%d%d",&n,&q);
        init();
        int tx,ty = n+1;
        for(int i = 1;i <= n;i ++){
            scanf("%d%d%lld",&a[i].u,&a[i].v,&a[i].val);
            if(vis[a[i].u]&&vis[a[i].v]){
                tx = a[i].u;
                a[i].u = n+1;
            }
            vis[a[i].u] = vis[a[i].v] = 1;
            add(a[i].u,a[i].v);
            add(a[i].v,a[i].u);
        }
       // cout<<"jsjd: "<<tx<<" "<<ty<<endl;
        dfs1(1,0,1);
        dfs2(1,1);
        for(int i = 1;i <= n;i ++){
            if(dep[a[i].u] < dep[a[i].v])swap(a[i].u,a[i].v);
            val[tid[a[i].u]] = a[i].val;
        }
        build(1,n+1,1);
      //  ct(1,n+1,1);cout<<endl;
        while(q--){
            int op,x,y;
            scanf("%d%d%d",&op,&x,&y);
            if(op == 0) update(tid[a[x].u],y,1,n+1,1);
            else{
                ll num = solve(x,tx)+solve(y,ty);
                ll num1 = solve(x,ty)+solve(y,tx);
                //cout<<"num: "<<num<<"num1 : "<<num1<<"solve "<<solve(x,y)<<endl;
                printf("%lld\n",min(solve(x,y),min(num,num1)));
            }
        }
    }
    return 0;
}      

转载于:https://www.cnblogs.com/kls123/p/9703786.html