TOJ 2609 Revamping Trails -- dp

題目：http://acm.tzc.edu.cn/acmhome/problemdetail.do?&method=showdetail&id=2609

分析：用dp[i][j]表示到結點i用了j次免費名額，那麼dp[i][j] = min(dp[t][j - 1], dp[t][j] + cost[t][j])，其中(t, i)之間有邊。一種做法就是像下面代碼中的那樣分k次調用dij，每次都固定k。另一種做法是隻調用一次dij，在考察結點t時，考慮t的每一個鄰接點i，考慮t與i之間用免費名額與不用兩種情況，也就是說對於每一條邊(t, i)，當考察(t, i)時，可能向隊列中加入兩個元素。但是這題比較卡時間，用後面這種做法會TLE。

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define mp make_pair
#define X first
#define Y second
#define MEMSET(a, b) memset(a, b, sizeof(a))
using namespace std;

typedef unsigned int ui;
typedef long long ll;
typedef unsigned long long ull;
typedef pair pii;
typedef vector vi;
typedef vi::iterator vi_it;
typedef map mii;
typedef priority_queue pqp;
typedef priority_queue, greater > rpqp;
typedef pair plli;
typedef priority_queue, greater > pq;

const int MAX_N = 10000 + 2;
const int MAX_M = 50000 + 2;
const int MAX_K = 20 + 2;
const ll INF = (ll)1.0e15;
int cnt = 1;
int head[MAX_N];
bool vis[MAX_N];
ll dis[MAX_N][MAX_K];
int level[MAX_N];

struct edge
{
	int v;
	int t;
	int next;
} eg[MAX_M << 1];

struct node
{
	int v;
	ll cost;
	
	node() {
	}
	
	node(int _v, const ll &_cost) : v(_v), cost(_cost) {
	}
	
	bool operator < (const node &nd) const {
		return cost > nd.cost;
	}
};

void add_edge(int u, int v, int w)
{
	eg[cnt].v = v;
	eg[cnt].t = w;
	eg[cnt].next = head[u];
	head[u] = cnt++;
	eg[cnt].v = u;
	eg[cnt].t = w;
	eg[cnt].next = head[v];
	head[v] = cnt++;
}

ll dij(int src, int des, int k)
{
	priority_queue q;
	dis[src][k] = 0;
	MEMSET(vis, 0);
	q.push(node(src, 0));
	if (!k) {
		while (!q.empty()) {
			node tmp = q.top();
			q.pop();
			int u = tmp.v;
			if (vis[u]) {
				continue;
			}
			vis[u] = true;
			for (int i = head[u]; i; i = eg[i].next) {
				int v = eg[i].v;
				if (!vis[v] && dis[u][0] + eg[i].t < dis[v][0]) {
					dis[v][0] = dis[u][0] + eg[i].t;
					q.push(node(v, dis[v][0]));
				}
			}
		}
	}
	else {								
		while (!q.empty()) {
			node tmp = q.top();
			q.pop();
			int u = tmp.v;
			if (vis[u]) {
				continue;
			}
			vis[u] = true;
			for (int i = head[u]; i; i = eg[i].next) {
				int v = eg[i].v;
				ll tt = min(dis[u][k - 1], dis[u][k] + eg[i].t);
				if (!vis[v] && tt < dis[v][k]) {
					dis[v][k] = tt;
					q.push(node(v, tt));
				}
			}
		}
	}
}

int main(int argc, char *argv[])
{
//	freopen("D:\\in.txt", "r", stdin);
	int n, m, k, i, j;
	cin >> n >> m >> k;
	while (m--) {
		int a, b, c;
		scanf("%d%d%d", &a, &b, &c);
		add_edge(a, b, c);
	}
	for (i = 1; i <= n; ++i) {
		for (j = 0; j <= k; ++j) {
			dis[i][j] = INF;
		}
	}
	for (i = 0; i <= k; ++i) {
		dij(1, n, i);
	}
	printf("%I64d\n", dis[n][k]);
	return 0;
}