UVa1048 Low Cost Air Travel--最短路

Posted 2021-11-27 dummyummy

很好的一道题呀

思路

状态\(d(i,j)\)表示已经经过了行程单中的\(i\)个城市，目前在城市\(j\)的最小代价，直接建边跑最短路就行了
比如机票为\(ACBD\)，行程单为\(CD\)，那么对于\((0,A)\)，连向\((1,C)\)，\((1,B)\)，\((2,D)\)
有两个需要注意的地方
1.起点为\((1,行程单的起点)\)
2.城市编号很大，要离散化
以下是代码，离散化用\(map\)完成

#include <algorithm>
#include  <iostream>
#include   <cstdlib>
#include   <cstring>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <cmath>
#include     <ctime>
#include     <queue>
#include       <map>
#include       <set>

using namespace std;

#define ull unsigned long long
#define pii pair<int, int>
#define uint unsigned int
#define mii map<int, int>
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define IINF 0x3f3f3f3f3f3f3f3fLL
#define DEF 0x8f8f8f8f
#define DDEF 0x8f8f8f8f8f8f8f8fLL
#define vi vector<int>
#define ll long long
#define mp make_pair
#define pb push_back
#define re register
#define il inline

#define N 10000

struct Edge {
  int next, from, to, w, id;
}e[2000000];

int ticketCnt, routeCnt, nodeCnt, cityCnt;
int price[250], cities[250];
vi tickets[250];
map<pii, int> nodeId;
mii cityId;
pii originNode[N+5];
int head[N+5], eid;
int d[N+5], pre[N+5];
bool inq[N+5];
int stk[N+5], tp;
queue<int> q;

void addEdge(int u, int v, int w, int id) {
  e[++eid] = Edge{head[u], u, v, w, id};
  head[u] = eid;
}

void spfa() {
  memset(d, 0x3f, sizeof d);
  memset(inq, 0, sizeof inq);
  memset(pre, 0, sizeof pre);
  int S = nodeId[mp(1, cities[1])];
  d[S] = 0;
  q.push(S);
  while(!q.empty()) {
    int u = q.front(); q.pop();
    inq[u] = 0;
    for(int i = head[u]; i; i = e[i].next) {
      int v = e[i].to, w = e[i].w;
      if(d[v] > d[u]+w) {
        d[v] = d[u]+w;
        pre[v] = i;
        if(!inq[v]) inq[v] = 1, q.push(v);
      }
    }
  }
}

void mark(int u) {
  if(!pre[u]) return ;
  stk[++tp] = e[pre[u]].id;
  mark(e[pre[u]].from);
}

int main() {
  int kase = 0;
  while(~scanf("%d", &ticketCnt) && ticketCnt) {
    ++kase;
    nodeCnt = cityCnt = 0;
    nodeId.clear();
    cityId.clear();
    for(int i = 1, cnt; i <= ticketCnt; ++i) {
      scanf("%d%d", &price[i], &cnt);
      tickets[i].clear();
      for(int j = 1, x; j <= cnt; ++j) {
        scanf("%d", &x);
        if(!cityId.count(x)) cityId[x] = ++cityCnt;
        tickets[i].pb(cityId[x]);
      }
    }
    scanf("%d", &routeCnt);
    for(int t = 1, len; t <= routeCnt; ++t) {
      memset(head, 0, sizeof head);
      eid = 0;
      scanf("%d", &len);
      for(int c = 1; c <= len; ++c) {
        scanf("%d", &cities[c]);
        if(!cityId.count(cities[c])) cityId[cities[c]] = ++cityCnt;
        cities[c] = cityId[cities[c]];
      }
      for(int ticket = 1; ticket <= ticketCnt; ++ticket) {
        for(int i = cities[1] == tickets[ticket][0]; i <= len; ++i) {
          int cnt = i;
          pii cur = mp(i, tickets[ticket][0]);
          if(!nodeId.count(cur)) nodeId[cur] = ++nodeCnt, originNode[nodeCnt] = cur;
          for(int j = 1; j < tickets[ticket].size(); ++j) {
            if(cnt+1 <= len && cities[cnt+1] == tickets[ticket][j]) cnt++;
            pii newState = mp(cnt, tickets[ticket][j]);
            if(!nodeId.count(newState)) nodeId[newState] = ++nodeCnt, originNode[nodeCnt] = newState;
            addEdge(nodeId[cur], nodeId[newState], price[ticket], ticket);
          }
        }
      }
      spfa();
      printf("Case %d, Trip %d: Cost = %d\n", kase, t, d[nodeId[mp(len, cities[len])]]);
      printf("  Tickets used: ");
      tp = 0;
      mark(nodeId[mp(len, cities[len])]);
      for(int i = tp; i > 1; --i) printf("%d ", stk[i]);
      printf("%d\n", stk[1]);
    }
  }
  return 0;
}