• 「USACO09FEB」改造路Revamping Trails


    传送门
    Luogu

    解题思路

    有点像这题,但是现在这道不能跑k遍SPFA了,会TLE。
    那么我们就跑分层图最短路,然后就变成模板题了。

    细节注意事项

    • 别跑SPFA就好了。

    参考代码

    #include <algorithm>
    #include <iostream>
    #include <cstring>
    #include <cstdio>
    #include <cctype>
    #include <cmath>
    #include <ctime>
    #include <queue>
    #define rg register
    using namespace std;
    template < class T > inline void read(T& s) {
    	s = 0; int f = 0; char c = getchar();
    	while (!isdigit(c)) f |= c == '-', c = getchar();
    	while (isdigit(c)) s = s * 10 + c - 48, c = getchar();
    	s = f ? -s : s;
    }
    
    typedef long long LL;
    const int _ = 210002;
    const int __ = 6100002;
    
    int tot, head[_], nxt[__], ver[__], w[__];
    inline void Add_edge(int u, int v, int d)
    { nxt[++tot] = head[u], head[u] = tot, ver[tot] = v, w[tot] = d; }
    
    int n, m, k, vis[_]; LL dis[_];
    
    inline int id(int u, int k) { return u + k * n; }
    
    inline void Dijkstra() {
    	static priority_queue < pair < LL, int > > Q;
    	memset(dis + 1, 0x3f, sizeof (LL) * (n * (k + 1)));
    	dis[id(1, 0)] = 0, Q.push(make_pair(0, id(1, 0)));
    	while (!Q.empty()) {
    		int u = Q.top().second; Q.pop();
    		if (vis[u]) continue; vis[u] = 1;
    		for (rg int i = head[u]; i; i = nxt[i]) {
    			int v = ver[i];
    			if (dis[v] > dis[u] + w[i])
    				dis[v] = dis[u] + w[i], Q.push(make_pair(-dis[v], v));
    		}
    	}
    }
    
    int main() {
    #ifndef ONLINE_JUDGE
    	freopen("cpp.in", "r", stdin);
    	freopen("cpp.out", "w", stdout);
    #endif
    	read(n), read(m), read(k);
    	for (rg int u, v, d; m--; ) {
    		read(u), read(v), read(d);
    		Add_edge(u, v, d), Add_edge(v, u, d);
    		for (rg int i = 1; i <= k; ++i) {
    			Add_edge(id(u, i), id(v, i), d);
    			Add_edge(id(v, i), id(u, i), d);
    			Add_edge(id(u, i - 1), id(v, i), 0);
    			Add_edge(id(v, i - 1), id(u, i), 0);
    		}
    	}
    	Dijkstra();
    	LL ans = 1e18;
    	for (rg int i = 0; i <= k; ++i)
    		ans = min(ans, dis[id(n, i)]);
    	printf("%lld
    ", ans);
    	return 0;
    }
    

    完结撒花 (qwq)

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  • 原文地址:https://www.cnblogs.com/zsbzsb/p/11805567.html
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