題目鏈接 BZOJ 2125
將原圖首先變成一棵樹，這裏就是對仙人掌進行圓方樹處理，於是接下去就是對新圖進行處理了。
比較簡單的，就是我們搜到的LCA點是圓點，那麼肯定此時的距離就是答案了，如果不是圓點呢？此時的LCA點是方點的話，那麼最後的最短距離還不好說，因爲存在可能走環的另一頭更近的可能性，或者它二者的距離可以更近，這裏的話，就是特別處理一下即可，是通過環上原距離，還是環的另外半圈更近，分別進行考慮即可。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define eps 1e-8
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
#define MAX_3(a, b, c) max(a, max(b, c))
#define Rabc(x) x > 0 ? x : -x
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 2e4 + 7, maxM = 1e5 + 7;
int N, M, Q;
struct Graph
{
int head[maxN], cnt;
struct Eddge
{
int nex, to; ll val;
Eddge(int a=-1, int b=0, ll c=0):nex(a), to(b), val(c) {}
} edge[maxM];
inline void addEddge(int u, int v, ll w) { edge[cnt] = Eddge(head[u], v, w); head[u] = cnt++; }
inline void _add(int u, int v, ll w) { addEddge(u, v, w); addEddge(v, u, w); }
inline void clear() { cnt = 0; for(int i=1; i<=N; i++) head[i] = -1; }
} Old, Now;
int dfn[maxN], low[maxN], tot, Stap[maxN], Stop, Bcnt;
ll dis[maxN], Tlen[maxN], las[maxN];
bool instack[maxN] = {false}, type[maxN];
vector<int> B_po[maxN];
inline void work(int u, int v)
{
Bcnt++; B_po[Bcnt].clear(); Now.head[N + Bcnt] = -1; Tlen[Bcnt] = dis[Stap[Stop]] - dis[u] + las[Stap[Stop]];
int p;
do
{
p = Stap[Stop--]; instack[p] = false;
ll d_1 = dis[p] - dis[u], d_2 = Tlen[Bcnt] - d_1;
type[p] = d_1 >= d_2;
Now._add(N + Bcnt, p, min(d_1, d_2));
B_po[Bcnt].push_back(p);
} while(p ^ v);
Now._add(u, N + Bcnt, 0);
}
void Tarjan(int u, int fa)
{
dfn[u] = low[u] = ++tot;
Stap[++Stop] = u;
instack[u] = true;
for(int i=Old.head[u], v; ~i; i=Old.edge[i].nex)
{
v = Old.edge[i].to;
if(v == fa) continue;
if(!dfn[v])
{
dis[v] = dis[u] + Old.edge[i].val;
Tarjan(v, u);
if(low[v] > dfn[u])
{
instack[Stap[Stop--]] = false;
Now._add(u, v, Old.edge[i].val);
}
else if(low[v] == dfn[u])
{
work(u, v);
}
low[u] = min(low[u], low[v]);
}
else if(instack[v])
{
if(low[u] > dfn[v])
{
low[u] = dfn[v];
las[u] = Old.edge[i].val;
}
}
}
}
int root[maxN][20], LOG_2[maxN], deep[maxN];
ll line[maxN][20];
void dfs(int u, int fa)
{
root[u][0] = fa; deep[u] = deep[fa] + 1;
for(int i=0; (1 << (i + 1)) < N + Bcnt; i++)
{
root[u][i + 1] = root[root[u][i]][i];
line[u][i + 1] = line[root[u][i]][i] + line[u][i];
}
for(int i=Now.head[u], v; ~i; i=Now.edge[i].nex)
{
v = Now.edge[i].to;
if(v == fa) continue;
line[v][0] = Now.edge[i].val;
dfs(v, u);
}
}
ll LCA(int u, int v)
{
ll ans = 0;
if(deep[u] < deep[v]) swap(u, v);
int det = deep[u] - deep[v];
for(int i=LOG_2[det]; i>=0; i--)
{
if((det >> i) & 1)
{
ans += line[u][i];
u = root[u][i];
}
}
if(u == v) return ans;
for(int i=LOG_2[N + Bcnt]; i>=0; i--)
{
if(root[u][i] ^ root[v][i])
{
ans += line[u][i] + line[v][i];
u = root[u][i];
v = root[v][i];
}
}
if(root[u][0] <= N)
{
ans += line[u][0] + line[v][0];
return ans;
}
ll Tdis = Tlen[root[u][0] - N];
if(type[u] == type[v])
{
ll d_1 = abs(dis[u] - dis[v]), d_2 = Tdis - d_1;
ans += min(d_1, d_2);
}
else
{
ll d_1 = line[u][0] + line[v][0], d_2 = Tdis - d_1;
ans += min(d_1, d_2);
}
return ans;
}
inline void init()
{
tot = Stop = Bcnt = 0;
Old.clear(); Now.clear();
for(int i = 1, j = 2, k = 0; i <= (N << 1); i++)
{
if(i == j) { k++; j <<= 1; }
LOG_2[i] = k;
}
}
int main()
{
scanf("%d%d%d", &N, &M, &Q);
init();
for(int i=1, u, v, w; i<=M; i++)
{
scanf("%d%d%d", &u, &v, &w);
Old._add(u, v, w);
}
dis[1] = 0;
Tarjan(1, 0);
deep[0] = 0;
dfs(1, 0);
int u, v;
while(Q --)
{
scanf("%d%d", &u, &v);
printf("%lld\n", LCA(u, v));
}
return 0;
}
/*
5 5 1
1 2 1
1 3 1
2 3 1
2 4 1
3 5 1
4 5
*/