[题解] 洛谷 P3393 逃离僵尸岛

题目TP门

很明显是一个最短路，但是如何建图才是关键。

对于每一个不可遍历到的点，可以向外扩散，找到危险城市。

若是对于每一个这样的城市进行搜索，时间复杂度就为\(O(n^2)\)，显然过不了。不妨把它们放在一个BFS里面进行搜索，先遍历可以向外延伸最长的点，因为这个点是可以存活很长的。若之后再遍历到这个点的时候，就不用在遍历了，因为这时候的存活时间已经没有之前遍历的时候高了。BFS+打标记遍历完一张图只需要\(O(n)\)的时间复杂度，很大程度上优化了建图方面。

最后根据所给的边跑一遍最短路即可，代码如下：

#include <queue>
#include <cstdio>
#include <vector>
#include <cstring>
using namespace std;
#define LL long long
#define INF 0x3f3f3f3f
void Quick_Read(LL &N) {
	N = 0;
	char c = getchar();
	LL op = 1;
	while(c < '0' || c > '9') {
		if(c == '-')
			op = -1;
		c = getchar();
	}
	while(c >= '0' && c <= '9') {
		N = (N << 1) + (N << 3) + c - 48;
		c = getchar();
	}
	N *= op;
}
const LL MAXN = 1e6 + 5;
struct Node {
	LL to, dist;
	Node() {}
	Node(LL T, LL D) {
		to = T;
		dist = D;
	}
	friend bool operator > (Node x, Node y) {
		return x.dist > y.dist;
	}
};
struct Flee {
	LL step, id;
	Flee() {}
	Flee(LL I, LL S) {
		step = S;
		id = I;
	}
	friend bool operator > (Flee x, Flee y) {
		return x.step < y.step;
	}
};
vector<LL> v[MAXN];
vector<Node> wg[MAXN];
queue<Flee> que;
priority_queue<Node, vector<Node>, greater<Node> > qu;
bool f[MAXN], vis[MAXN], vi[MAXN], V[MAXN];
LL c[MAXN], w[MAXN];
LL A[MAXN], B[MAXN], d[MAXN];
LL n, m, k, s, p, q;
void Dijkstra() {
	memset(d, 0x3f, sizeof(d));
	qu.push(Node(1, 0));
	d[1] = 0;
	while(!qu.empty()) {
		LL now = qu.top().to; qu.pop();
		if(V[now])
			continue;
		V[now] = true;
		LL SIZ = wg[now].size();
		for(LL i = 0; i < SIZ; i++) {
			LL next = wg[now][i].to;
			if(d[now] + wg[now][i].dist < d[next]) {
				d[next] = d[now] + wg[now][i].dist;
				qu.push(Node(next, d[next]));
			}
		}
	}
	printf("%lld", d[n]);
}
void bfs() {
	for(LL i = 1; i <= k; i++) {
		que.push(Flee(c[i], s));
		vi[c[i]] = 1;
	}
	while(!que.empty()) {
		Flee now = que.front(); que.pop();
		if(vis[now.id])
			continue;
		vis[now.id] = 1;
		vi[now.id] = 1;
		if(now.step == 0)
			continue;
		LL SIZ = v[now.id].size();
		for(LL i = 0; i < SIZ; i++) {
			LL nex = v[now.id][i];
			que.push(Flee(nex, now.step - 1));
		}
	}
}
void Read() {
	Quick_Read(n);
	Quick_Read(m);
	Quick_Read(k);
	Quick_Read(s);
	Quick_Read(p);
	Quick_Read(q);
	for(LL i = 1; i <= k; i++) {
		Quick_Read(c[i]);
		f[c[i]] = true;
	}
	for(LL i = 1; i <= m; i++) {
		Quick_Read(A[i]);
		Quick_Read(B[i]);
		v[A[i]].push_back(B[i]);
		v[B[i]].push_back(A[i]);
	}
}
void Build() {
	for(LL i = 1; i <= n; i++)
		w[i] = p;
	for(LL i = 1; i <= n; i++)
		if(vi[i])
			w[i] = q;
	for(LL i = 1; i <= k; i++)
		w[c[i]] = INF;
	for(LL i = 1; i <= m; i++) {
		if(B[i] == n) {
			wg[A[i]].push_back(Node(B[i], 0));
			wg[B[i]].push_back(Node(A[i], w[A[i]]));
		}
		else if(A[i] == n) {
			wg[A[i]].push_back(Node(B[i], w[B[i]]));
			wg[B[i]].push_back(Node(A[i], 0));
		}
		else {
			wg[A[i]].push_back(Node(B[i], w[B[i]]));
			wg[B[i]].push_back(Node(A[i], w[A[i]]));
		}
	}
}
int main() {
	Read();
	bfs();
	Build();
	Dijkstra();
	return 0;
}