PTA 森森旅游 (30 分) | 堆优化迪杰斯特拉

Posted 2022-04-19 PushyTao

6 11 3
1 2 3 5
1 3 8 4
2 4 4 6
3 1 8 6
1 3 10 8
2 3 2 8
3 4 5 3
3 5 10 7
3 3 2 3
4 6 10 12
5 6 10 6
3 4 5 2 5 100
1 2
2 1
1 17

8
8
1

Code:

typedef pair<long long, int> PII;
int n, m, q;
int head1[maxn], head2[maxn];
ll dis1[maxn], dis2[maxn];
struct node 
	int v, to, nex;
	int w;
e[maxn << 2];
int idx;
multiset<ll> mst;
void init() 
	idx = 0;
	memset(head1, -1, sizeof head1);
	memset(head2, -1, sizeof head2);

void add(int h[], int u, int v, int w) 
	e[idx].v = v;
	e[idx].nex = h[u];
	e[idx].w = w;
	h[u] = idx++;

bool vis[maxn];
void dij(ll dis[], int head[], int st) 
	memset(dis, 0x3f3f3f3f3f3f3f3f, sizeof dis1);
	memset(vis, 0, sizeof vis);
	priority_queue <PII, vector<PII>, greater<PII> > que;
	que.push( 0,st );
	dis[st] = 0;
	while (que.size()) 
		int u = que.top().second;
		que.pop();
		if (vis[u]) continue;
		vis[u] = 1;
		for (int i = head[u]; ~i; i = e[i].nex) 
			//	cout << e[i].v << endl;
			int to = e[i].v;
			if (dis[to] > dis[u] + e[i].w) 
				dis[to] = dis[u] + e[i].w;
				que.push( dis[to],to );
			
		
	

int a[maxn];
void getAns() 
	for (int i = 1; i <= n; i++) 
		if (dis1[i] == 0x3f3f3f3f3f3f3f3f) continue;
		if (dis2[i] == 0x3f3f3f3f3f3f3f3f) continue;
		ll val = dis1[i] + (dis2[i] + a[i] - 1) / a[i];
		mst.insert(val);
		// puts("add");
	
	for (int qq = 1; qq <= q; qq++) 
		// 	puts("qq");
		int x = read, aa = read;
		if (dis1[x] != 0x3f3f3f3f3f3f3f3f && dis2[x] != 0x3f3f3f3f3f3f3f3f) 
			ll t = dis1[x] + (dis2[x] + a[x] - 1) / a[x];
			mst.erase(mst.find(t));
			a[x] = aa;
			ll val = dis1[x] + (dis2[x] + aa - 1) / aa;
			mst.insert(val);
		
		printf("%lld\\n", *mst.begin());
	

int main() 
	n = read, m = read, q = read;
	init();
	memset(dis1, 0x3f3f3f3f3f3f3f3f, sizeof dis1);
	memset(dis2, 0x3f3f3f3f3f3f3f3f, sizeof dis2);
	for (int i = 1; i <= m; i++) 
		int u = read, v = read, c = read, d = read;
		add(head1, u, v, c);
		add(head2, v, u, d);
	
	for (int i = 1; i <= n; i++) a[i] = read;
	dij(dis1, head1, 1);
	dij(dis2, head2, n);
	/*
	for (int i = 1; i <= n; i++) 
		cout << dis1[i] << " " << dis2[i] << endl;
	*/
	getAns();
	return 0;

/**
6 11 3
1 2 3 5
1 3 8 4
2 4 4 6
3 1 8 6
1 3 10 8
2 3 2 8
3 4 5 3
3 5 10 7
3 3 2 3
4 6 10 12
5 6 10 6
3 4 5 2 5 100
1 2
2 1
1 17

**/