Hangzhou Electric Multi-school seventh field traffic Network in Numazu

Problem Description

Chika is elected mayor of Numazu. She needs to manage the traffic. To manage the traffic was too hard for her. So she needs your help.
You were given the map of the City--an undirected connected weighted graph with n nodes and n edges, and you had to fini SH Q missions. Each mission consists of 3 integers OP, X and Y.
When op=0, you need to modify the weight of the Xth edge to Y.
When op=1, you need to calculate the length of the shortest path from node X to node Y.

Input

The first line contains a single integer T, the number of test cases.
Each test case starts with a line containing the integers N and Q, the number of nodes (and edges) and the number of Queri Es. (3≤n≤105) (1≤q≤105)
Each of the following N lines contain the description of the edges. The ith line represents the ith edge, which contains 3 space-separated integers UI, vi, and WI. This means the there is an undirected edge between nodes UI and VI, with a weight of wi. (1≤ui,vi≤n) (1≤wi≤105)
Then Q lines follow, the ith line contains 3 integers OP, X and Y. The meaning has been described above. (0≤op≤1) (1≤x≤105) (1≤y≤105)
It is guaranteed, the graph contains no self loops or multiple edges.

Output

For each test case, and for each mission whose op=1, print one line containing one integer, the length of the shortest pat H between X and Y.

Sample Input

2
5 5
1 2 3
2 3 5
2 4 5
2 5 1
4 3 3
0 1 5
1 3 2
1 5 4
0 5 4
1 5 1
5 3
1 2 3
1 3 2
3 4 4
4 5 5
2 5 5
0 1 3
0 4 1
1 1 4

Sample Output

5
6
6
6

Analysis

When the game is silly, change the tree to a base ring tree.

• Remove any edge of the ring first, and two points of the side connection are recorded as R1,r2
• For each query, the distance between x and Y is three: (1) The shortest distance from the tree after the edge is removed, and (2) x->r1->r2->y; (3) x->r2->r1->y. Take a minimum of three values to

Code

#include <iostream> #include <cstdio> #include <algorithm> #include <cstring> #include < queue> #include <map> #include <cassert> #include <cmath> #include <vector>using namespace Std;typedef Long Long ll;const int maxn=100050;int fa[maxn];int Find (int x) {return x==fa[x]?x: (Fa[x]=find (fa[x]));} struct Edge {int v,nxt,d;}    E[maxn*2];int h[maxn],tot,n,q;void addedge (int x,int y,int d) {e[++tot]= (Edge) {y,h[x],d}; H[x]=tot;}    ll sum[maxn];void Add (int x,int val) {assert (x>0);        while (x<=n) {sum[x]+=val;    x+= (x&-x);    }}LL query (int x) {ll ans=0;        while (x) {ans+=sum[x];    x-= (x&-x); } return ans;    int f[maxn][20],idx[maxn],cnt,dep[maxn],r[maxn];void dfs (int x,int par,int val) {fa[x]=f[x][0]=par;    dep[x]=dep[par]+1;    idx[x]=++cnt;    Add (Cnt,val);    for (int i = 1; f[x][i-1]; ++i) f[x][i]=f[f[x][i-1]][i-1]; for (int i = h[x]; i; i=e[i].nxt) {if (PAR!=E[I].V) DFS (E[I].V,X,E[I].D);    } r[x]=cnt; Add (cnt+1,-val);}    int LCA (int x,int y) {if (Dep[x]<dep[y]) swap (x, y);    int h=dep[x]-dep[y];    for (int i = n; i >= 0; i) {if (h& (1<<i)) x=f[x][i];    } if (x==y) return x;    for (int i = n; i >= 0; i.) {if (F[x][i]!=f[y][i]) x=f[x][i],y=f[y][i]; } return f[x][0];}    ll dis (int x,int y) {int Z=lca (x, y); return query (Idx[x]) +query (Idx[y])-query (idx[z]) * *;}    int X[maxn],y[maxn],w[maxn];int Main () {int T;    scanf ("%d", &t);        while (t--) {scanf ("%d%d", &n,&q);        for (int i = 0; I <= N; ++i) h[i]=0,fa[i]=i,sum[i]=0;        cnt=0,tot=0;        int r1=0,r2=0,bw=0,id;            for (int i = 1; I <= n; ++i) {int x,y,w;            scanf ("%d%d%d", &x,&y,&w);            if (find (x) ==find (y)) {r1=x,r2=y,bw=w,id=i;            } else Addedge (x,y,w), Addedge (y,x,w), Fa[find (x)]=find (y);        X[i]=x,y[i]=y,w[i]=w;     }   DFS (1,0,0);            while (q--) {int op,x,y;            scanf ("%d%d%d", &op,&x,&y);                if (op==0) {if (id==x) bw=y; else {int u= (fa[x[x]]==y[x]?                    X[X]:Y[X]);                    Add (idx[u],-w[x]+y);                    Add (r[u]+1,w[x]-y);                W[x]=y;                }} else {ll ans=dis (x, y);                Ans=min (Ans,dis (r1,x) +dis (r2,y) +BW);                Ans=min (Ans,dis (r1,y) +dis (r2,x) +BW);            printf ("%lld\n", ans); }}} return 0;}

Hangzhou Electric Multi-school seventh field traffic Network in Numazu