题目链接CF-786B
题意:有N个点,Q次操作,图的起点是S,问经过Q次操作之后,S到每个点的最短路。
- 增加从点v到点u的距离w的边
- 增加从点v到点[l, r]区间的每个点的距离w的边
- 增加从区间[l, r]到点v的距离为w的边
于是,我们可以开两棵线段树,分别表示出入的情况,然后操作结束之后跑一次堆优化的Dijkstra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 0x3f3f3f3f
#define BIG_INF 0x3f3f3f3f3f3f3f3f
#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
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
const int maxN = 1e5 + 7;
int N, Q, S, tot, rt1, rt2, head[maxN << 3], 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[maxN << 5];
inline void addEddge(int u, int v, ll w)
{
edge[cnt] = Eddge(head[u], v, w);
head[u] = cnt++;
}
namespace Segement
{
inline int new_node() { ++tot; head[tot] = -1; return tot; }
int lc[maxN << 3], rc[maxN << 3];
void build(int &u, int fa, int l, int r, int op)
{
u = new_node();
if(fa)
{
if(op) addEddge(u, fa, 0);
else addEddge(fa, u, 0);
}
if(l == r) return;
int mid = HalF;
build(lc[u], u, l, mid, op);
build(rc[u], u, mid + 1, r, op);
}
inline int point_query(int rt, int l, int r, int qx)
{
if(l == r) return rt;
int mid = HalF;
if(qx <= mid) return point_query(lc[rt], l, mid, qx);
else return point_query(rc[rt], mid + 1, r, qx);
}
inline void Range_query(int rt, int l, int r, int ql, int qr, int u, int w, int op)
{
if(ql <= l && qr >= r)
{
if(op) addEddge(u, rt, w);
else addEddge(rt, u, w);
return;
}
int mid = HalF;
if(qr <= mid) Range_query(lc[rt], l, mid, ql, qr, u, w, op);
else if(ql > mid) Range_query(rc[rt], mid + 1, r, ql, qr, u, w, op);
else { Range_query(lc[rt], l, mid, ql, qr, u, w, op); Range_query(rc[rt], mid + 1, r, ql, qr, u, w, op); }
}
};
using namespace Segement;
namespace Dijkstra
{
ll dis[maxN << 3];
struct node
{
int id; ll val;
node(int a=0, ll b=0):id(a), val(b) {}
friend bool operator < (node e1, node e2) { return e1.val > e2.val; }
};
priority_queue<node> Que;
void Dij()
{
for(int i=1; i<=tot; i++) dis[i] = BIG_INF;
int tmp = point_query(rt1, 1, N, S);
dis[tmp] = 0;
Que.push(node(tmp, 0));
while(!Que.empty())
{
node now = Que.top(); Que.pop();
int u = now.id;
if(dis[u] < now.val) continue;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(dis[v] > dis[u] + edge[i].val)
{
dis[v] = dis[u] + edge[i].val;
Que.push(node(v, dis[v]));
}
}
}
}
};
using namespace Dijkstra;
inline void init()
{
cnt = tot = 0;
build(rt1, 0, 1, N, 0);
build(rt2, 0, 1, N, 1);
for(int i=1; i<=N; i++) addEddge(point_query(rt1, 1, N, i), point_query(rt2, 1, N, i), 0);
}
int main()
{
scanf("%d%d%d", &N, &Q, &S);
init();
for(int i=1, op, v, l, r, w; i<=Q; i++)
{
scanf("%d", &op);
switch (op)
{
case 1:
{
scanf("%d%d%d", &v, &l, &w);
addEddge(point_query(rt2, 1, N, v), point_query(rt1, 1, N, l), w);
break;
}
case 2:
{
scanf("%d%d%d%d", &v, &l, &r, &w);
Range_query(rt1, 1, N, l, r, point_query(rt2, 1, N, v), w, 1);
break;
}
default:
{
scanf("%d%d%d%d", &v, &l, &r, &w);
Range_query(rt2, 1, N, l, r, point_query(rt1, 1, N, v), w, 0);
break;
}
}
}
Dij();
ll ans;
for(int i=1; i<=N; i++)
{
ans = dis[point_query(rt2, 1, N, i)];
printf("%lld%c", ans == BIG_INF ? -1 : ans, i == N ? '\n' : ' ');
}
return 0;
}