2019牛客暑期多校训练营（第四场）J-free

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题意：给你n个城市，m条道路，经过每一条要花费这条路的代价，现给你k个机会，使得最多k条路的代价为0，问从起点s到终点t花费的最少代价

思路：分层图最短路经典裸题

方法一

Code

#include <bits/stdc++.h>
using namespace std;
struct edge { int to, cost; };
typedef pair<int, int> P; // first是最短距离，second是顶点的编号 

const int MAX_V = 2000005;
int n, m, s, t, k;
int d[MAX_V];
vector<edge> G[MAX_V];

void dijkstra()
{
    priority_queue<P, vector<P>, greater<P> >que;
    memset(d, 0x3f, sizeof(d));
    d[s] = 0;
    que.push(P(0, s));
    
    while (!que.empty()) {
        P p = que.top(); que.pop();
        int v = p.second;
        if (d[v] < p.first) continue;
        for (int i = 0; i < G[v].size(); i++) {
            edge e = G[v][i];
            if (d[e.to] > d[v] + e.cost) {
                d[e.to] = d[v] + e.cost;
                que.push(P(d[e.to], e.to));
            }
        }
    }
}

int main()
{
    scanf("%d%d%d%d%d", &n, &m, &s, &t, &k);
    for(int i = 1; i <= m; i++) {
        int u, v, cost;
        scanf("%d%d%d", &u, &v, &cost);
        for(int j = 0; j <= k; j++) {
            G[u+n*j].push_back({v+n*j,cost});
            G[v+n*j].push_back({u+n*j,cost});
            if(j < k) {
                G[u+n*j].push_back({v+n*(j+1),0});
                G[v+n*j].push_back({u+n*(j+1),0});
            }
        }
    }
    dijkstra();
    //int ans=0x3f3f3f3f;
    //for(int i=0;i<=k;i++) ans=min(ans,dis[t+n*i]);
    printf("%d\n", d[t+k*n]);
    return 0;
}

View Code

方法二

让我最惊讶的是这题有的人的Dijkstra算法没经过堆优化直接用暴力出来？？？后来一看n，m都小于1e3，复杂度最多也就1e6，还爆不了，好吧。当然也有用优化过的算法或者是SPFA，明天再过来看一下

Code

#include <bits/stdc++.h>
using namespace std;
typedef pair<int, int> P;

const int MAX_N=1005;
int n, m, s, t, k;
vector<P> G[MAX_N];
int d[MAX_N][MAX_N];

void dijkstra()
{
    memset(d, 0x3f, sizeof(d));
    d[s][0]=0;
    queue<P> que; 
    que.push({s,0});
    
    while(!que.empty()) {
        int u = que.front().first, t = que.front().second; que.pop();
        for(int i = 0; i < G[u].size(); i++) {
            int v = G[u][i].first, cost = G[u][i].second;
            if(d[v][t]>d[u][t]+cost) d[v][t] = d[u][t]+cost, que.push({v,t});
            if(t<k&&d[v][t+1]>d[u][t]) d[v][t+1] = d[u][t], que.push({v,t+1});
        }
    }
}

int main()
{
    scanf("%d%d%d%d%d", &n, &m, &s, &t, &k);
    for(int i=0;i<m;i++) {
        int u, v, cost;
        scanf("%d%d%d", &u, &v, &cost);
        G[u].push_back({v,cost});
        G[v].push_back({u,cost});
    }
    dijkstra();
    printf("%d\n",d[t][k]);
    return 0;
}

View Code