2018icpc南京网络赛-L Magical Girl Haze (分层图最短路)

题意：

有向图，可以把k条路的长度变为0，求1到n的最短路

思路：

将图复制k份，一共k+1层图，对于每一条i→j,都连一条低层的i→高层的j，并且权值为0

即对每一对<i,j,w>，都加边<i,j,w>,<i+n,j+n,w>,<i+2n,j+2n,w>,....,<i+kn,j+kn,w>

同时加“楼梯”<i,j+n,0>,<i+n,j+2n,0>,...,<i+(k-1)n, j+kn>

然后跑一个1~(k+1)n的最短路，用迪杰斯特拉+堆优化

代码：

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cmath>
#include<cstring>
#include<string>
#include<stack>
#include<queue>
#include<deque>
#include<set>
#include<vector>
#include<map>
#include<functional>

#define fst first
#define sc second
#define pb push_back
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,mid,root<<1
#define rson mid+1,r,root<<1|1
#define lc root<<1
#define rc root<<1|1
#define lowbit(x) ((x)&(-x)) 

using namespace std;

typedef double db;
typedef long double ldb;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> PI;
typedef pair<ll,ll> PLL;

const db eps = 1e-;
const int mod = 1e9+;
const int maxn = 2e6+;
const int maxm = 2e6+;
const int inf = 0x3f3f3f3f;
const db pi = acos(-1.0);

int dist[maxn];
struct node{
    int id, d;
    node(){}
    node(int a,int b) {id = a; d = b;}
    bool operator < (const node & a)const{
        if(d == a.d) return id > a.id;
        else return d > a.d;
    }
};
vector<node>e[maxn];
int n;

void dijkstra(int s){
    for(int i = ; i <= n; i++) dist[i] = inf;
    dist[s] = ;
    priority_queue<node>q;
    q.push(node(s, dist[s]));
    while(!q.empty()){
        node top = q.top();
        q.pop();
        if(top.d != dist[top.id]) continue;
        for(int i = ; i < (int)e[top.id].size(); i++){
            node x = e[top.id][i];
            if(dist[x.id] > top.d + x.d){
                dist[x.id] = top.d + x.d;
                q.push(node(x.id, dist[x.id]));
            }
        }
    }
    return;
}

int main() {
    int T;
    scanf("%d", &T);
    while(T--){
        int m, k;
        scanf("%d %d %d", &n, &m, &k);
        while(m--){
            int x ,y, w;
            scanf("%d %d %d", &x, &y, &w);
            for(int i = ; i <= k; i++){
                e[x + i*n].pb(node(y + i*n, w));
                if(i)e[x + (i-)*n].pb(node(y + i*n, ));

            }
        }
        n += k*n;
        dijkstra();
        printf("%d\n", dist[n]);
        for(int i = ; i <= n; i++)e[i].clear();
    }
    return ;
}