題意

有三種邊[u,v], [u, [L,R]], [[L,R], v],求從S出發(fā)到其他所有點的最短路

思路

線段樹維護(hù)區(qū)間建圖

#include <bits/stdc++.h> const int maxn = 1e5 + 5; const long long inf = 1e18; const int mod = 1e9 + 7; using namespace std; int T[maxn<<2], num[2][maxn<<2], cnt; struct ac{long long v, c;bool operator <(const ac &t) const{return t.c < c;} }; vector<ac> g[maxn<<2]; void add(int u, int v, int c) {g[u].push_back({v, c}); } int build(int rt, int l, int r, int op) {if (l == r) return num[op][rt] = l;num[op][rt] = ++cnt;int mid = (l + r) >> 1;int L = build(rt<<1, l, mid, op);int R = build(rt<<1|1, mid+1, r, op);if (op == 0) {add(L, num[op][rt], 0);add(R, num[op][rt], 0);}else {add(num[op][rt], L, 0);add(num[op][rt], R, 0);}return num[op][rt]; } void update(int rt, int l, int r, int u, int ql, int qr, int c, int op) {if (ql > r || qr < l) return;if (l >= ql && r <= qr) {if (op == 0) add(u, num[1][rt], c);else add(num[0][rt], u, c);return; }int mid = (l + r) >> 1;update(rt<<1, l, mid, u, ql, qr, c, op);update(rt<<1|1, mid+1, r, u, ql, qr, c, op); } long long dis[maxn<<3]; bool vis[maxn<<3]; void Dijkstra(int s, int n) {fill(dis, dis+n+1, inf);fill(vis, vis+n+1, 0);dis[s] = 0;priority_queue<ac> que;que.push({s, 0});while (!que.empty()) {ac f = que.top();que.pop();int u = f.v;if (dis[u] < f.c || vis[u]) continue;vis[u] = 1;for (int i = 0; i < (int)g[u].size(); ++i) {int v = g[u][i].v;long long c = f.c + g[u][i].c;if (dis[v] > c) {dis[v] = c;que.push({v, c});}}} } int main() {int n, q, s;scanf("%d %d %d", &n, &q, &s);cnt = n;build(1, 1, n, 0);build(1, 1, n, 1);for (int i = 0, u,ql,qr,c,op; i < q; ++i) {scanf("%d%d%d", &op, &u, &ql);if (op > 1) scanf("%d%d", &qr, &c);else scanf("%d", &c), qr = ql;if (op == 1) add(u, ql, c);else update(1, 1, n, u, ql, qr, c, op-2);}Dijkstra(s, cnt);for (int i = 1; i <= n; ++i) {if (i > 1) printf(" ");printf("%lld", dis[i]==inf?-1:dis[i]);}puts("");return 0; }

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