我们观察题目后发现这很明显是一道有关最短路的题，首先可以无脑打一个最短路模板上去。
36分
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define pr pair<int,int>
using namespace std;
int n, m, w, h, tot;
const int N = 70010, M = 150000;
int head[N], dis[N], vis[N], to[M], nt[M], val[M];
priority_queue<pr >q;
struct dian {int x, y;} d[N];
struct tiao {int p, t, l, r, d, u;} t[M];
inline int read()
{
int res = 0; char ch = getchar(); bool XX = false;
for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
return XX ? -res : res;
}
void add(int f, int t, int d)
{
to[++tot] = t; val[tot] = d; nt[tot] = head[f]; head[f] = tot;
}
void DIJ()
{
int x;
memset(dis, 0x3f, sizeof(dis));
q.push(pr(0, 1)); dis[1] = 0;
while (!q.empty())
{
x = q.top().second; q.pop();
if (vis[x])continue; vis[x] = 1;
for (int i = head[x]; i; i = nt[i])
if (dis[to[i]] > dis[x] + val[i])
{
dis[to[i]] = dis[x] + val[i];
q.push(pr(-dis[to[i]], to[i]));
}
}
}
int main()
{
cin >> n >> m >> w >> h;
for (int i = 1; i <= n; ++i)d[i].x = read(), d[i].y = read();
for (int i = 1; i <= m; ++i)
{
t[i].p = read(), t[i].t = read(), t[i].l = read(), t[i].r = read(), t[i].d = read(), t[i].u = read();
for (int j = 1; j <= n; ++j)
if (j != t[i].p && t[i].l <= d[j].x && d[j].x <= t[i].r && t[i].d <= d[j].y && d[j].y <= t[i].u)add(t[i].p, j, t[i].t);
}
DIJ();
for (int i = 2; i <= n; ++i)printf("%d\n", dis[i]);
return 0;
}
思考对于 \(h=1\) 的数据，我们可以用线段树优化建图水过。
那 \(h！=1\) 的话，我们用二维数据结构优化建图就行，这里我用的是 \(KDtree\)
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define lson ch[k][0]
#define rson ch[k][1]
#define pr pair<int,int>
using namespace std;
int n, m, w, h, tot, num, root, nowk;
const int N = 140010, M = 3000010;
int head[N], dis[N], vis[N], to[M], nt[M], val[M], s[N], ch[N][2], minx[N], miny[N], maxx[N], maxy[N];
priority_queue<pr >q;
struct dian {int x, y, id;} d[N];
struct tiao {int p, t, l, r, d, u;} t[M];
inline int read()
{
int res = 0; char ch = getchar(); bool XX = false;
for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
return XX ? -res : res;
}
void add(int f, int t, int d)
{
to[++tot] = t; val[tot] = d; nt[tot] = head[f]; head[f] = tot;
}
void pushup(int k)
{
minx[k] = maxx[k] = d[k - n].x; miny[k] = maxy[k] = d[k - n].y; add(k, k - n, 0);
if (lson)minx[k] = min(minx[k], minx[lson]), maxx[k] = max(maxx[k], maxx[lson]), miny[k] = min(miny[k], miny[lson]), maxy[k] = max(maxy[k], maxy[lson]), add(k, lson, 0);
if (rson)minx[k] = min(minx[k], minx[rson]), maxx[k] = max(maxx[k], maxx[rson]), miny[k] = min(miny[k], miny[rson]), maxy[k] = max(maxy[k], maxy[rson]), add(k, rson, 0);
}
int my(int a, int b) {return nowk ? d[a].x < d[b].x : d[a].y < d[b].y;}
void build(int &k, int l, int r, int d)
{
if (l > r)return;
int mid = (l + r) >> 1; nowk = d;
nth_element(s + l, s + mid, s + r + 1, my); k = s[mid] + n;
build(lson, l, mid - 1, k ^ 1); build(rson, mid + 1, r, k ^ 1);
pushup(k);
}
void ADD(int k, int now)
{
if (maxx[k] < t[now].l || t[now].r < minx[k] || maxy[k] < t[now].d || t[now].u < miny[k] )return;
if (t[now].l <= minx[k] && maxx[k] <= t[now].r && t[now].d <= miny[k] && maxy[k] <= t[now].u) {add(t[now].p, k, t[now].t); return;}
if (t[now].l <= d[k - n].x && d[k - n].x <= t[now].r && t[now].d <= d[k - n].y && d[k - n].y <= t[now].u)add(t[now].p, k - n, t[now].t);
if (lson)ADD(lson, now); if (rson)ADD(rson, now);
}
void DIJ()
{
int x;
memset(dis, 0x3f, sizeof(dis));
q.push(pr(0, 1)); dis[1] = 0;
while (!q.empty())
{
x = q.top().second; q.pop();
if (vis[x])continue; vis[x] = 1;
for (int i = head[x]; i; i = nt[i])
if (dis[to[i]] > dis[x] + val[i])
{
dis[to[i]] = dis[x] + val[i];
q.push(pr(-dis[to[i]], to[i]));
}
}
}
int main()
{
cin >> n >> m >> w >> h;
for (int i = 1; i <= n; ++i)d[i].x = read(), d[i].y = read(), d[i].id = i, s[i] = i;
build(root, 1, n, 0);
for (int i = 1; i <= m; ++i)
{
t[i].p = read(), t[i].t = read(), t[i].l = read(), t[i].r = read(), t[i].d = read(), t[i].u = read();
ADD(root, i);
}
DIJ();
for (int i = 2; i <= n; ++i)printf("%d\n", dis[i]);
return 0;
}
被卡空间.jpg。边数太多了，我们无法把边全都建出来，又因为 \(Dij\) 的特性，一个点只会遍历一次和它相连的边，所以我们只需要用到边的时候再去找就行。
根据实现60~80不等
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#define lson ch[k][0]
#define rson ch[k][1]
#define pr pair<int,int>
using namespace std;
int n, m, w, h, num, root, nowk, now;
const int N = 140010, M = 150010;
int dis[N], vis[N], s[N], ch[N][2], minx[N], miny[N], maxx[N], maxy[N];
priority_queue<pr >q;
vector<int>v[N];
struct dian {int x, y;} d[N];
struct tiao {int p, t, l, r, d, u;} t[M];
inline int read()
{
int res = 0; char ch = getchar(); bool XX = false;
for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
return XX ? -res : res;
}
inline void pushup(int k)
{
minx[k] = maxx[k] = d[k - n].x; miny[k] = maxy[k] = d[k - n].y;
if (lson)minx[k] = min(minx[k], minx[lson]), maxx[k] = max(maxx[k], maxx[lson]), miny[k] = min(miny[k], miny[lson]), maxy[k] = max(maxy[k], maxy[lson]);
if (rson)minx[k] = min(minx[k], minx[rson]), maxx[k] = max(maxx[k], maxx[rson]), miny[k] = min(miny[k], miny[rson]), maxy[k] = max(maxy[k], maxy[rson]);
}
inline int my(int a, int b) {return nowk ? d[a].x < d[b].x : d[a].y < d[b].y;}
void build(int &k, int l, int r, int d)
{
if (l > r)return;
int mid = (l + r) >> 1; nowk = d;
nth_element(s + l, s + mid, s + r + 1, my); k = s[mid] + n;
build(lson, l, mid - 1, k ^ 1); build(rson, mid + 1, r, k ^ 1);
pushup(k);
}
inline void geng(int y, int val)
{
if (dis[y] > val)dis[y] = val, q.push(pr(-dis[y], y));
}
void ADD(int k)
{
if (vis[k])return;
if (maxx[k] < t[now].l || t[now].r < minx[k] || maxy[k] < t[now].d || t[now].u < miny[k] )return;
if (t[now].l <= minx[k] && maxx[k] <= t[now].r && t[now].d <= miny[k] && maxy[k] <= t[now].u) {geng(k, dis[t[now].p] + t[now].t); return;}
if (t[now].l <= d[k - n].x && d[k - n].x <= t[now].r && t[now].d <= d[k - n].y && d[k - n].y <= t[now].u)geng(k - n, dis[t[now].p] + t[now].t);
if (lson)ADD(lson); if (rson)ADD(rson);
}
inline void DIJ()
{
int x;
memset(dis, 0x3f, sizeof(dis));
q.push(pr(0, 1)); dis[1] = 0;
while (!q.empty())
{
x = q.top().second; q.pop();
if (vis[x])continue; vis[x] = 1;
if (x <= n)
{
for (int i = 0, siz = v[x].size(); i < siz; ++i)now = v[x][i], ADD(root);
}
else
{
if (ch[x][0])geng(ch[x][0], dis[x]);
if (ch[x][1])geng(ch[x][1], dis[x]);
geng(x - n, dis[x]);
}
}
}
signed main()
{
cin >> n >> m >> w >> h;
for (int i = 1; i <= n; ++i)d[i].x = read(), d[i].y = read(), s[i] = i;
build(root, 1, n, 0);
for (int i = 1; i <= m; ++i)
t[i].p = read(), t[i].t = read(), t[i].l = read(), t[i].r = read(), t[i].d = read(), t[i].u = read(), v[t[i].p].push_back(i);
DIJ();
for (int i = 2; i <= n; ++i)printf("%d\n", dis[i]);
return 0;
}
我们发现 \(Dij\) 每次是 \(1\) .取出dis最小的点 \(2\) .更新和它相连的点 \(3\).删去该点。
这些 \(KDtree\) 本身都可以完成，直接在 \(KDtree\) 上维护就行了。
\(shu\) 里的变量
\(p,val,id,vis\) ：该点对应的点的坐标， \(dis\) ，编号，是否取出过
\(l,r,mn,mx,tag\) :区间的坐标最大/小值， \(dis\) 的最大/小值，对区间打的标记。
100分
#include<algorithm>
#include<iostream>
#include<cstdio>
#include<vector>
#define lson tr[k].ls
#define rson tr[k].rs
using namespace std;
int n, m, w, h, p, t, l, r, d, u, root, nowk, cnt;
const int N = 70010, inf = 0x3f3f3f3f;
int s[N], dis[N];
struct dian {int x, y;} di[N];
struct bian {dian l, r; int dis;};
struct shu {dian p, l, r; int ls, rs, mn, mx, val, id, tag, vis;} tr[N];
vector<bian>v[N];
inline int read()
{
int res = 0; char ch = getchar(); bool XX = false;
for (; !isdigit(ch); ch = getchar())(ch == '-') && (XX = true);
for (; isdigit(ch); ch = getchar())res = (res << 3) + (res << 1) + (ch ^ 48);
return XX ? -res : res;
}
void pushup(int k)
{
if (tr[k].vis)tr[k].l = (dian) {inf, inf}, tr[k].r = (dian) { -inf, -inf};
else tr[k].l = tr[k].r = tr[k].p;
if (lson)tr[k].l.x = min(tr[k].l.x, tr[lson].l.x), tr[k].l.y = min(tr[k].l.y, tr[lson].l.y), tr[k].r.x = max(tr[k].r.x, tr[lson].r.x), tr[k].r.y = max(tr[k].r.y, tr[lson].r.y);
if (rson)tr[k].l.x = min(tr[k].l.x, tr[rson].l.x), tr[k].l.y = min(tr[k].l.y, tr[rson].l.y), tr[k].r.x = max(tr[k].r.x, tr[rson].r.x), tr[k].r.y = max(tr[k].r.y, tr[rson].r.y);
}
void upd(int k)
{
if (tr[k].vis)tr[k].mn = inf, tr[k].mx = -inf;
else tr[k].mn = tr[k].mx = tr[k].val;
if (lson)tr[k].mn = min(tr[k].mn, tr[lson].mn), tr[k].mx = max(tr[k].mx, tr[lson].mx);
if (rson)tr[k].mn = min(tr[k].mn, tr[rson].mn), tr[k].mx = max(tr[k].mx, tr[rson].mx);
}
int my(int a, int b)
{
return nowk ? di[a].x < di[b].x : di[a].y < di[b].y;
}
void build(int &k, int l, int r, int now)
{
if (l > r)return; k = ++cnt; nowk = now;
int mid = (l + r) >> 1;
nth_element(s + l, s + mid, s + r + 1, my);
tr[k].id = s[mid]; tr[k].p = di[s[mid]]; tr[k].val = tr[k].tag = inf;
if (tr[k].id == 1)tr[k].val = 0;
build(lson, l, mid - 1, now ^ 1); build(rson, mid + 1, r, now ^ 1);
pushup(k); upd(k);
}
void work(int k, int v)
{
if (v < tr[k].mx && v < tr[k].tag)
{
tr[k].mx = tr[k].tag = v; tr[k].mn = min(tr[k].mn, v);
if (!tr[k].vis)tr[k].val = min(tr[k].val, v);
}
}
void pushdown(int k)
{
if (tr[k].tag == inf)return;
if (lson)work(lson, tr[k].tag); if (rson)work(rson, tr[k].tag);
tr[k].tag = inf;
}
int find(int k, int v)
{
int res;
pushdown(k);
if (!tr[k].vis && v == tr[k].val) {tr[k].vis = 1; pushup(k); upd(k); return k;}
if (lson && tr[lson].mn == v)res = find(lson, v);
else res = find(rson, v);
pushup(k); upd(k);
return res;
}
void change(int k, int v, dian l, dian r)
{
pushdown(k);
if (v >= tr[k].mx)return;
if (r.x < tr[k].l.x || tr[k].r.x < l.x || r.y < tr[k].l.y || tr[k].r.y < l.y)return;
if (l.x <= tr[k].l.x && tr[k].r.x <= r.x && l.y <= tr[k].l.y && tr[k].r.y <= r.y) {work(k, v); return;}
if (!tr[k].vis && l.x <= tr[k].p.x && tr[k].p.x <= r.x && l.y <= tr[k].p.y && tr[k].p.y <= r.y)tr[k].val = min(tr[k].val, v);
if (lson)change(lson, v, l, r); if (rson)change(rson, v, l, r);
pushup(k); upd(k);
}
signed main()
{
cin >> n >> m >> w >> h;
for (int i = 1; i <= n; ++i)di[i].x = read(), di[i].y = read(), s[i] = i;
for (int i = 1; i <= m; ++i)
{
p = read(), t = read(), l = read(), r = read(), d = read(), u = read();
v[p].push_back((bian) {(dian) {l, d}, (dian) {r, u}, t});
}
build(root, 1, n, 0);
for (int i = 1, pos, x; i <= n; ++i)
{
x = tr[pos = find(root, tr[root].mn)].id; dis[x] = tr[pos].val;
for (int j = 0, siz = v[x].size(); j < siz; ++j)change(root, dis[x] + v[x][j].dis, v[x][j].l, v[x][j].r);
}
for (int i = 2; i <= n; ++i)printf("%d\n", dis[i]);
return 0;
}
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