Miraclys

一言(ヒトコト)

洛谷P4568 [JLOI2011]飞行路线

\(\Large\textbf{Solution: } \large{容易想到跑分层图。就是细节比较多,数组一定要开大!!!}\)

\(\Large\textbf{Code: }\)

#include <bits/stdc++.h>
#define gc() getchar()
using namespace std;
const int N = 220005;
const int inf = 0x7fffffff;
int n, m, s, t, k, cnt, ret, head[N], dis[N], vis[N];

struct Edge {
	int to, next, val;	
}e[50005 * 45];

struct Node {
	int num, d;
	friend bool operator < (Node a, Node b) {
		return a.d > b.d;
	}
};
priority_queue<Node> q;

inline int read() {
	int x = 0;
	char ch = gc();
	while (!isdigit(ch)) ch = gc();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = gc();
	return x; 
}

inline void add(int x, int y, int val) {
	e[++cnt].to = y;
	e[cnt].val = val;
	e[cnt].next = head[x];
	head[x] = cnt;
}

inline void Dijkstra() {
	for (int i = 1; i <=(k + 1) * n; ++i) dis[i] = inf;
	dis[s] = 0;
	q.push((Node){s, 0});
	while (!q.empty()) {
		Node cur = q.top(); q.pop();
		int fr = cur.num;
		if (vis[fr]) continue;
		vis[fr] = 1;
		for (int i = head[fr]; i ; i = e[i].next) {
			int u = e[i].to;
			if (vis[u]) continue;
			if (dis[u] > dis[fr] + e[i].val) dis[u] = dis[fr] + e[i].val, q.push((Node) {u, dis[u]});
		}
	}
	return;
}

int main() {
	n = read(), m = read(), k = read();
	s = read(), t = read();
	++s, ++t;
	int x, y, w;
	for (int i = 1; i <= m; ++i) {
		x = read(), y = read(), w = read(); ++x, ++y;
		add(x, y, w), add(y, x, w);
		for (int j = 1; j <= k; ++j) {
			add(x + (j - 1) * n, y + j * n, 0); add(y + (j - 1) * n, x + j * n, 0);
			add(x + j * n, y + j * n, w); add(y + j * n, x + j * n, w);
		}
	}
	for (int i = 1; i <= k; ++i) add(n * (i - 1) + t, n * i + t, 0);
	Dijkstra();
	printf("%d\n", dis[n * k + t]); 
	return 0;
} 
posted @ 2020-03-24 22:11  Miraclys  阅读(94)  评论(0编辑  收藏  举报

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BNDong

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