Blog of RuSun

\begin {array}{c} \mathfrak {One Problem Is Difficult} \\\\ \mathfrak {Because You Don't Know} \\\\ \mathfrak {Why It Is Diffucult} \end {array}

CF786B Legacy

LuoGu: CF786B Legacy

CF: B. Legacy

麻烦的是要维护两课线段树。

查看代码
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#include <cstdio>
#include <queue>
using namespace std;
typedef long long LL;
typedef pair<LL, int> PLI;
const int N = 1e5 + 10, M = 1e7 + 10;
const LL inf = 1e14 + 1;
int idx, hd[N << 3], nxt[M], edg[M];
LL d[N << 3], wt[M];
int n, m, st, id[N];
bool vis[N << 3];
void add(int a, int b, LL c)
{
nxt[++idx] = hd[a];
hd[a] = idx;
edg[idx] = b;
wt[idx] = c;
}
struct Node
{
int l, r;
};
struct Seg
{
Node tr[N << 2];
void build(int x, int l, int r, bool op)
{
tr[x].l = l;
tr[x].r = r;
if (l == r)
{
id[l] = x;
return;
}
int mid = l + r >> 1;
build(x << 1, l, mid, op);
build(x << 1 | 1, mid + 1, r, op);
if (op)
{
add((x << 1) + (n << 2), x + (n << 2), 0);
add((x << 1 | 1) + (n << 2), x + (n << 2), 0);
}
else
{
add(x, x << 1, 0);
add(x, x << 1 | 1, 0);
}
}
void addRange(int x, int pos, int l, int r, LL w, bool op)
{
if (tr[x].l >= l && tr[x].r <= r)
{
if (op)
add(x + (n << 2), id[pos], w);
else
add(id[pos] + (n << 2), x, w);
return;
}
int mid = tr[x].l + tr[x].r >> 1;
if (l <= mid)
addRange(x << 1, pos, l, r, w, op);
if (r > mid)
addRange(x << 1 | 1, pos, l, r, w, op);
}
} A, B;
int main()
{
scanf("%d%d%d", &n, &m, &st);
for (int i = 1; i <= (n << 3); i++)
hd[i] = -1;
A.build(1, 1, n, 0);
B.build(1, 1, n, 1);
for (int i = 1; i <= n; i++)
add(id[i], id[i] + (n << 2), 0);
for (int op; m; m--)
{
scanf("%d", &op);
if (op == 1)
{
int u, v;
LL w;
scanf("%d%d%lld", &u, &v, &w);
add(id[u], id[v], w);
}
else if (op == 2)
{
int u, l, r;
LL w;
scanf("%d%d%d%lld", &u, &l, &r, &w);
A.addRange(1, u, l, r, w, 0);
}
else if (op == 3)
{
int u, l, r;
LL w;
scanf("%d%d%d%lld", &u, &l, &r, &w);
B.addRange(1, u, l, r, w, 1);
}
}
for (int i = 1; i <= (n << 3); i++)
d[i] = inf;
d[id[st]] = 0;
priority_queue<PLI, vector<PLI>, greater<PLI>> q;
q.push(make_pair(0, id[st]));
while (!q.empty())
{
int t = q.top().second;
q.pop();
if (vis[t])
continue;
vis[t] = true;
for (int i = hd[t]; ~i; i = nxt[i])
if (d[t] + wt[i] < d[edg[i]])
{
d[edg[i]] = d[t] + wt[i];
q.push(make_pair(d[edg[i]], edg[i]));
}
}
for (int i = 1; i <= n; i++)
if (d[id[i]] == inf)
printf("-1 ");
else
printf("%lld ", d[id[i]]);
return 0;
}