傳送門
恰逢才做過VFK的A+B Problem,發現這道題也可以那樣搞。區間連邊的時候,我們就可以給那個區間在線段樹對應的標號上連邊。
線段樹也可以不建出來,直接當做一個標號的合集,不佔用內存,只用模擬在線段樹上找區間的過程就可以了。
如果不清楚的話,可以看下A+B Problem的題解裏面配有插圖。
連好邊之後就可以直接跑
分析一下時間複雜度,線段樹節點數是
#include <cstdio>
#include <cstring>
#define MAXN 200005
#define MAXM 5000005
#pragma comment(linker, "/STACK:102400000,102400000")
#define INF 0x3f3f3f3f3f3f3f3fLL
#define I64 long long
struct { int v, nxt, w; } e[MAXM << 1];
int Adj[MAXN * 6], c, n, L[MAXN], R[MAXN], N;
I64 C[MAXN];
inline void Add(int u, int v, I64 w) { ++ c; e[c].v = v; e[c].nxt = Adj[u]; e[c].w = w; Adj[u] = c; }
inline void GET(int &n) {
static char c; n = 0;
do c = getchar(); while('0' > c || c > '9');
do (n*=10)+=c-'0',c=getchar(); while('0' <= c && c <= '9');
}
inline void GET(I64 &n) {
static char c; n = 0;
do c = getchar(); while('0' > c || c > '9');
do (n*=10)+=c-'0',c=getchar(); while('0' <= c && c <= '9');
}
int LL, RR, F;
void Link(int i, int l, int r) {
if(LL > r || l > RR) return;
if(LL <= l && r <= RR) { Add(F, n + i, C[F]); return; }
int mid = (l + r) >> 1;
Link(i << 1, l, mid);
Link(i<<1|1, mid+1, r);
}
void Build(int i, int l, int r) {
if(l == r) { Add(n + i, l, 0); if(i + n > N) N = i+n; return; }
Add(i + n, n + (i << 1), 0); Add(n + i, n + (i << 1|1), 0);
int mid = (l + r) >> 1;
Build(i<<1, l, mid); Build(i<<1|1, mid+1, r);
}
/***************************************/
I64 dis[MAXN * 6];
int minp[MAXN * 12];
bool used[MAXN * 6];
inline void pushup(int p) {
int &r = minp[p];
r = p * !used[p];
if (dis[minp[p<<1]] < dis[r]) r = minp[p<<1];
if (dis[minp[p<<1|1]] < dis[r]) r = minp[p<<1|1];
}
void relax(int p, I64 d) {
if(d >= dis[p]) return;
dis[p] = d;
for(int i = p; i; i>>=1) pushup(i);
}
void finish(int p) {
used[p] = 1;
for(int i = p; i; i>>=1) pushup(i);
}
/***************以上這一段可以就當做堆優化**************/
void Dijkstra() {
for(int i = 0; i <= N; ++ i)
dis[i] = INF, used[i] = (minp[i] = 0);
relax(1, 0);
while(minp[1]) {
int u = minp[1];
finish(u);
for(int i = Adj[u]; i; i = e[i].nxt)
relax(e[i].v, dis[u] + (I64)e[i].w);
}
}
int main() {
int T; scanf("%d", &T);
while(T --) {
GET(n); c = N = 0; memset(Adj, 0, sizeof Adj);
for(int i = 1; i <= n; ++ i) GET(L[i]);
for(int i = 1; i <= n; ++ i) GET(R[i]);
for(int i = 1; i <= n; ++ i) GET(C[i]);
Build(1, 1, n);
for(int i = 1; i <= n; ++ i) {
LL = L[i]+i; RR = R[i]+i; F = i;
Link(1, 1, n);
RR = i-L[i]; LL = i-R[i];
Link(1, 1, n);
}
Dijkstra();
putchar('0');
for(int i = 2; i <= n; ++ i) printf(" %I64d", dis[i] == INF ? -1 : dis[i]);
puts("");
}
return 0;
}