题目链接：https://codeforces.ml/contest/1163/problem/F
题目大意：给你一个n个点m条边的无向图。q次修改，每次修改编号为x的边权值为y，每次修改后输出1-n的的最短路。
#include <bits/stdc++.h>
#define pii pair<int, int>
#define LL long long
#define piii pair<pii, int>
#define ls(x) (x << 1)
#define rs(x) ((x << 1) | 1)
using namespace std;
const int maxn = 200010;
vector<piii> G[maxn];
piii edge[maxn];
int n;
map<int, int> mp;
void add(int x, int y, int z, int id) {
G[x].push_back(make_pair(make_pair(y, z), id));
G[y].push_back(make_pair(make_pair(x, z), id));
}
struct node {
bool flag;
LL mi, Set;
};
struct SegmentTree {
int n;
node tr[maxn * 4];
void pushup(int o) {
tr[o].mi = min(tr[ls(o)].mi, tr[rs(o)].mi);
}
void maintain(int o, LL val) {
tr[o].Set = min(tr[o].Set, val);
tr[o].mi = min(tr[o].mi, tr[o].Set);
tr[o].flag = 1;
}
void pushdown(int o) {
if(tr[o].flag) {
maintain(ls(o), tr[o].Set);
maintain(rs(o), tr[o].Set);
tr[o].Set = 1e18;
tr[o].flag = 0;
}
}
void build(int o, int l, int r) {
if(l == r) {
tr[o].mi = 1e18;
tr[o].flag = 0;
tr[o].Set = 1e18;
return;
}
int mid = (l + r) >> 1;
build(ls(o), l, mid);
build(rs(o), mid + 1, r);
tr[o].mi = 1e18;
tr[o].flag = 0;
tr[o].Set = 1e18;
}
void update(int o, int l, int r, int ql, int qr, LL val) {
if(ql > qr) return;
if(l == 0) return;
if(l >= ql && r <= qr) {
maintain(o, val);
return;
}
pushdown(o);
int mid = (l + r) >> 1;
if(ql <= mid) update(ls(o), l, mid, ql, qr, val);
if(qr > mid) update(rs(o), mid + 1, r, ql, qr, val);
pushup(o);
}
LL query(int o, int l, int r, int pos) {
if(l == r && l == pos) {
return tr[o].mi;
}
pushdown(o);
int mid = (l + r) >> 1;
if(pos <= mid) return query(ls(o), l, mid, pos);
else return query(rs(o), mid + 1, r, pos);
}
};
SegmentTree st;
struct Dj {
priority_queue<pair<long long, int> > q;
pii pre[maxn];
bool in_line[maxn], v[maxn], in_tree[maxn], is_line[maxn];
LL dis[maxn];
vector<int> G1[maxn];
int f[maxn];
vector<int> line;
vector<LL> re;
void add1(int x, int y) {
G1[x].push_back(y);
G1[y].push_back(x);
}
void dijkstra(int s) {
memset(v, 0, sizeof(v));
memset(dis, 0x3f, sizeof(dis));
q.push(make_pair(0, s));
dis[s] = 0;
while(q.size()) {
int x = q.top().second;
q.pop();
if(v[x]) continue;
v[x] = 1;
for (int i = 0; i < G[x].size(); i++) {
int y = G[x][i].first.first;//to
LL z = G[x][i].first.second;//w
if(v[y]) continue;
if(dis[y] > dis[x] + z) {
dis[y] = dis[x] + z;
pre[y] = make_pair(x, G[x][i].second);//由x节点来，边的编号为G[x][i].second
q.push(make_pair(-dis[y], y));
}
}
}
}
void dfs(int x, int flag, int fa) {
f[x] = flag;
for (int i = 0 ; i < G1[x].size(); i++) {
int y = G1[x][i];
if(y == fa || is_line[y]) continue;
dfs(y, flag, x);
}
}
void solve(int s) {
for (int i = 1; i <= n; i++) {
if(i == s) continue;
add1(i, pre[i].first);
in_tree[pre[i].second] = 1;//编号为pre[i].second在树上
}
for (int i = n + 1 - s; i != s; i = pre[i].first) {
line.push_back(i);
in_line[pre[i].second] = 1;//编号为re[i].second在线段树上
is_line[i] = 1;//节点i在线段树上面
}
line.push_back(s);
is_line[s] = 1;
for (int i = 0; i < line.size(); i++) {
int y = line[i];
dfs(y, y, -1);
// cout<<y<<" f[] = "<<'\n';
// for(int i=1; i<=n; i++){
// printf("%d - %d\n", i, f[i]);
// }
// printf("*******************\n");
}
}
};
Dj dj1, dj2;
int main() {
int x, y, z, m, T;
// freopen("1163F.in", "r", stdin);
// freopen("1163F.out", "w", stdout);
scanf("%d%d%d", &n, &m, &T);
for (int i = 1; i <= m; i++) {
scanf("%d%d%d", &x, &y, &z);
edge[i] = make_pair(make_pair(x, y), z);
add(x, y, z, i);
}
dj1.dijkstra(1), dj2.dijkstra(n);
dj1.solve(1), dj2.solve(n);
int cnt = 0;
for (int i = dj1.line.size() - 1; i >= 0; i--) {
mp[dj1.line[i]] = ++cnt;//对最短短路径上的点编号建树
}
st.build(1, 1, cnt - 1);
for (int i = 1; i <= m; i++) {
if(dj1.in_tree[i] && dj2.in_tree[i]) continue;
else {
int x = edge[i].first.first, y = edge[i].first.second;
LL tmp = 1e18;
int l, r;
l = min(mp[dj1.f[x]], mp[dj2.f[y]]), r = max(mp[dj1.f[x]], mp[dj2.f[y]]);
tmp = dj1.dis[x] + dj2.dis[y] + edge[i].second;
//cout<<"id: "<<i<<" "<<x<<"->"<<y<<" ["<<l<<","<<r-1<<"]"<<endl;
if(l >= 1 && r <= cnt)
st.update(1, 1, cnt - 1, l, r - 1, tmp);
swap(x, y);
l = min(mp[dj1.f[x]], mp[dj2.f[y]]), r = max(mp[dj1.f[x]], mp[dj2.f[y]]);
//cout<<"id: "<<i<<" "<<x<<"->"<<y<<" ["<<l<<","<<r-1<<"]"<<endl;
tmp = dj1.dis[x] + dj2.dis[y] + edge[i].second;
if(l >= 1 && r <= cnt)
st.update(1, 1, cnt - 1, l, r - 1, tmp);
}
}
int cnt_T = 0;
while(T--) {
cnt_T++;
LL ans = 0;
scanf("%d%d", &x, &y);
int l1 = edge[x].first.first, r1 = edge[x].first.second;
//
if(!dj1.in_line[x]) {//如果边没有在线段树上
ans = dj1.dis[n];
ans = min(ans, min(dj1.dis[l1] + dj2.dis[r1] + y, dj1.dis[r1] + dj2.dis[l1] + y));
printf("%lld\n", ans);
} else {
LL ans = dj1.dis[l1] + dj2.dis[r1] + y;
ans = min(ans, dj1.dis[r1] + dj2.dis[l1] + y);
int now = min(mp[l1], mp[r1]);
printf("%lld\n", min(ans, st.query(1, 1, cnt - 1, now)));
}
}
}
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