[题解] [JSOI2010] 旅行

题面

题解

发现数据范围很小， 考虑从这上面入手
不难发现， 如果我们把所有边的长度排序， 将每条边选与不选看作一个 01 串
假设最优路径长度为 L , 必然存在一个 (K) , 满足前 (1 o K) 都是 1 ， 其他的随便
考虑枚举这个 (K)
设 (f[i][j][k]) 满足到 (i) 点， 前 (K) 个中选了 (j) 条， 已经交换了 (k) 条
转移的话就是看这条边是不是可以换， 换不换就行
这里的转移方程和网上大部分的不太一样， 我把前 (K) 条的贡献提前加上去了
发现可以最短路转移， 跑就对了

Code

#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
const int N = 55;
const int M = 155; 
using namespace std;

int head[N], cnt, n, m, K, sum[M], f[N][M][25], ans, U[M], V[M], W[M]; 
struct edge { int to, nxt, cost, id; } e[M << 1];
struct pir
{
	int a, b;
	pir(int _a = 0, int _b = 0) { a = _a, b = _b; }
	bool operator < (const pir &p) const
		{
			return a < p.a; 
		}
} c[M << 1]; 
struct node
{
	int a, b, c, d;
	node(int _a = 0, int _b = 0, int _c = 0, int _d = 0) { a = _a, b = _b, c = _c, d = _d; }
	bool operator < (const node &p) const
		{
			return d > p.d; 
		}
};
bool vis[N][M][25]; 
priority_queue<node> q; 

template < typename T >
inline T read()
{
	T x = 0, w = 1; char c = getchar();
	while(c < '0' || c > '9') { if(c == '-') w = -1; c = getchar(); }
	while(c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar(); }
	return x * w; 
}

inline void adde(int u, int v, int w, int id) { e[++cnt] = (edge) { v, head[u], w, id }, head[u] = cnt; }

void dijkstra(int lim)
{
	for(int i = 1; i <= n; i++)
		for(int j = 0; j <= lim; j++)
			for(int k = 0; k <= K; k++)
				f[i][j][k] = 0x3f3f3f3f, vis[i][j][k] = 0; 
	f[1][0][0] = sum[lim]; q.push(node(1, 0, 0, f[1][0][0])); 
	int u, j, k, w; 
	while(!q.empty())
	{
		node tmp = q.top(); q.pop(); 
		u = tmp.a, j = tmp.b, k = tmp.c, w = tmp.d; 
		if(vis[u][j][k]) continue; 
		vis[u][j][k] = 1; 
		for(int v, i = head[u]; i; i = e[i].nxt)
		{
			v = e[i].to; 
			if(e[i].id <= lim)
			{
				if(j + 1 <= lim && w < f[v][j + 1][k])
				{
					f[v][j + 1][k] = w;
					q.push(node(v, j + 1, k, f[v][j + 1][k])); 
				}
			}
			else
			{
				if(w + e[i].cost < f[v][j][k])
					f[v][j][k] = w + e[i].cost, q.push(node(v, j, k, f[v][j][k])); 
				if(j < lim && k < K && w < f[v][j + 1][k + 1])
				   f[v][j + 1][k + 1] = w, q.push(node(v, j + 1, k + 1, f[v][j + 1][k + 1])); 
			}
		}
	}
	for(int i = 0; i <= K; i++)
		ans = min(ans, f[n][lim][i]); 
}

int main()
{
	n = read <int> (), m = read <int> (), K = read <int> ();
	ans = 0x3f3f3f3f; 
	for(int i = 1; i <= m; i++)
		U[i] = read <int> (), V[i] = read <int> (), W[i] = read <int> (), c[i] = pir(W[i], i);
	sort(c + 1, c + m + 1);
	for(int i = 1; i <= m; i++)
		adde(U[c[i].b], V[c[i].b], W[c[i].b], i), adde(V[c[i].b], U[c[i].b], W[c[i].b], i); 
	for(int i = 1; i <= m; i++)
		sum[i] = sum[i - 1] + c[i].a; 
	for(int i = 1; i <= m; i++)
		dijkstra(i); 
	printf("%d
", ans); 
	return 0; 
}