[USACO09FEB]改造路Revamping Trails 分层最短路 Dijkstra BZOJ 1579
题意翻译
约翰一共有N)个牧场.由M条布满尘埃的小径连接.小径可 以双向通行.每天早上约翰从牧场1出发到牧场N去给奶牛检查身体.
通过每条小径都需要消耗一定的时间.约翰打算升级其中K条小径，使之成为高 速公路.在高速公路上的通行几乎是瞬间完成的，所以高速公路的通行时间为0.
请帮助约翰决定对哪些小径进行升级，使他每天从1号牧场到第N号牧场所花的时间最短
题目描述
Farmer John dutifully checks on the cows every day. He traverses some of the M (1 <= M <= 50,000) trails conveniently numbered 1..M from pasture 1 all the way out to pasture N (a journey which is always possible for trail maps given in the test data). The N (1 <= N <= 10,000) pastures conveniently numbered 1..N on Farmer John's farm are currently connected by bidirectional dirt trails. Each trail i connects pastures P1_i and P2_i (1 <= P1_i <= N; 1 <= P2_i <= N) and requires T_i (1 <= T_i <= 1,000,000) units of time to traverse.
He wants to revamp some of the trails on his farm to save time on his long journey. Specifically, he will choose K (1 <= K <= 20) trails to turn into highways, which will effectively reduce the trail's traversal time to 0. Help FJ decide which trails to revamp to minimize the resulting time of getting from pasture 1 to N.
TIME LIMIT: 2 seconds
输入输出格式
输入格式：* Line 1: Three space-separated integers: N, M, and K
* Lines 2..M+1: Line i+1 describes trail i with three space-separated integers: P1_i, P2_i, and T_i
输出格式：* Line 1: The length of the shortest path after revamping no more than K edges
输入输出样例
说明
K is 1; revamp trail 3->4 to take time 0 instead of 100. The new shortest path is 1->3->4, total traversal time now 1.
比较裸的题目，与之相似的是：[JLOI2011]飞行路线
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
//#include<cctype>
//#pragma GCC optimize("O3")
using namespace std;
#define maxn 1000005
#define inf 0x3f3f3f3f
#define INF 9999999999
#define rdint(x) scanf("%d",&x)
#define rdllt(x) scanf("%lld",&x)
#define rdult(x) scanf("%lu",&x)
#define rdlf(x) scanf("%lf",&x)
#define rdstr(x) scanf("%s",x)
typedef long long ll;
typedef unsigned long long ull;
typedef unsigned int U;
#define ms(x) memset((x),0,sizeof(x))
const long long int mod = 1e9 + 7;
#define Mod 1000000000
#define sq(x) (x)*(x)
#define eps 1e-3
typedef pair<int, int> pii;
#define pi acos(-1.0)
//const int N = 1005;
#define REP(i,n) for(int i=0;i<(n);i++)
typedef pair<int, int> pii;
inline ll rd() {
ll x = 0;
char c = getchar();
bool f = false;
while (!isdigit(c)) {
if (c == '-') f = true;
c = getchar();
}
while (isdigit(c)) {
x = (x << 1) + (x << 3) + (c ^ 48);
c = getchar();
}
return f ? -x : x;
}
ll gcd(ll a, ll b) {
return b == 0 ? a : gcd(b, a%b);
}
ll sqr(ll x) { return x * x; }
/*ll ans;
ll exgcd(ll a, ll b, ll &x, ll &y) {
if (!b) {
x = 1; y = 0; return a;
}
ans = exgcd(b, a%b, x, y);
ll t = x; x = y; y = t - a / b * y;
return ans;
}
*/
ll qpow(ll a, ll b, ll c) {
ll ans = 1;
a = a % c;
while (b) {
if (b % 2)ans = ans * a%c;
b /= 2; a = a * a%c;
}
return ans;
}
int n, m, k;
int d[maxn][30];
int vis[maxn][30];
vector<pii>vc[maxn];
int s, t;
struct node {
int v, lev, dis;
node(){}
node(int v,int lev,int dis):v(v),lev(lev),dis(dis){}
bool operator < (const node&rhs)const {
return dis > rhs.dis;
}
};
priority_queue<node>q;
int main()
{
//ios::sync_with_stdio(0);
rdint(n); rdint(m); rdint(k);
for (int i = 1; i <= m; i++) {
int u, v, w;
rdint(u); rdint(v); rdint(w);
vc[u].push_back(make_pair(v, w));
vc[v].push_back(make_pair(u, w));
}
for (int i = 1; i <= n; i++) {
for (int j = 0; j <= k; j++)d[i][j] = inf;
}
d[1][0] = 0;
q.push(node(1, 0, 0));
while (!q.empty()) {
node tmp = q.top(); q.pop();
int to = tmp.v, lev = tmp.lev;
if (vis[to][lev])continue;
vis[to][lev] = 1;
vector<pii>::iterator it;
for (it = vc[to].begin(); it != vc[to].end(); it++) {
int y = (*it).first, w = (*it).second;
if (d[y][lev] > d[to][lev] + w) {
d[y][lev] = d[to][lev] + w;
q.push(node(y, lev, d[y][lev]));
}
if (lev < k) {
if (d[y][lev + 1] > d[to][lev]) {
d[y][lev + 1] = d[to][lev];
q.push(node(y, lev + 1, d[y][lev + 1]));
}
}
}
}
int ans = d[n][0];
for (int i = 1; i <= k; i++) {
ans = min(ans, d[n][i]);
}
cout << ans << endl;
return 0;
}
