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  • UVa 11367 Full Tank? (DP + Dijkstra)

    题意:n个城市有m条道路。每个城市的油价不一样,给出起点s和终点t,以及汽车的油箱的容量,求从城市s到城市 t 的最便宜路径。

    析:dp[u][i] 表示在第 u 个城市,还剩下 i L升油,一开始用BFS,TLE,要注意效率,用dijkstra,找到城市 t 就该结束了。

    代码如下:

    #pragma comment(linker, "/STACK:1024000000,1024000000")
    #include <cstdio>
    #include <string>
    #include <cstdlib>
    #include <cmath>
    #include <iostream>
    #include <cstring>
    #include <set>
    #include <queue>
    #include <algorithm>
    #include <vector>
    #include <map>
    #include <cctype>
    #include <cmath>
    #include <stack>
    #include <sstream>
    #include <list>
    #include <assert.h>
    #include <bitset>
    #define debug() puts("++++");
    #define gcd(a, b) __gcd(a, b)
    #define lson l,m,rt<<1
    #define rson m+1,r,rt<<1|1
    #define fi first
    #define se second
    #define pb push_back
    #define sqr(x) ((x)*(x))
    #define ms(a,b) memset(a, b, sizeof a)
    #define sz size()
    #define pu push_up
    #define pd push_down
    #define cl clear()
    #define all 1,n,1
    #define FOR(i,x,n)  for(int i = (x); i < (n); ++i)
    #define freopenr freopen("in.txt", "r", stdin)
    #define freopenw freopen("out.txt", "w", stdout)
    using namespace std;
    
    typedef long long LL;
    typedef unsigned long long ULL;
    typedef pair<int, int> P;
    const int INF = 0x3f3f3f3f;
    const double inf = 1e20;
    const double PI = acos(-1.0);
    const double eps = 1e-8;
    const int maxn = 1000 + 10;
    const int maxm = 1e5 + 10;
    const int mod = 50007;
    const int dr[] = {-1, 0, 1, 0};
    const int dc[] = {0, -1, 0, 1};
    const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
    int n, m;
    const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
    inline bool is_in(int r, int c) {
      return r >= 0 && r < n && c >= 0 && c < m;
    }
    
    int pri[maxn];
    struct Edge{
      int to, val, next;
    };
    Edge edges[maxn*10<<1];
    int head[maxn], cnt;
    
    void addEdge(int u, int v, int c){
      edges[cnt].to = v;
      edges[cnt].val = c;
      edges[cnt].next = head[u];
      head[u] = cnt++;
    }
    
    int dp[maxn][105];
    
    struct HeapNode{
      int cost, u, last;
      bool operator < (const HeapNode &p) const{
        return cost > p.cost;
      }
    };
    
    int main(){
      while(scanf("%d %d", &n, &m) == 2){
        for(int i = 0; i < n; ++i)  scanf("%d", pri + i);
        ms(head, -1);  cnt = 0;
        while(m--){
          int u, v, c;
          scanf("%d %d %d", &u, &v, &c);
          addEdge(u, v, c);
          addEdge(v, u, c);
        }
        scanf("%d", &m);
        while(m--){
          int s, t, c;
          scanf("%d %d %d", &c, &s, &t);
          for(int i = 0; i < n; ++i)  ms(dp[i], INF);
          dp[s][0] = 0;
          priority_queue<HeapNode> pq;
          pq.push((HeapNode){0, s, 0});
          bool ok = false;
          while(!pq.empty()){
            HeapNode h = pq.top();  pq.pop();
            if(h.u == t){ printf("%d
    ", h.cost);  ok = true;  break; }
            int u = h.u, last = h.last, cost = h.cost;
            if(last < c && dp[u][last+1] > cost + pri[u]){
              dp[u][last+1] = cost + pri[u];
              pq.push((HeapNode){cost + pri[u], u, last + 1});
            }
            for(int i = head[u]; ~i; i = edges[i].next){
              int v = edges[i].to;
              if(last >= edges[i].val && dp[v][last-edges[i].val] > cost){
                dp[v][last-edges[i].val] = cost;
                pq.push((HeapNode){cost, v, last-edges[i].val});
              }
            }
          }
          if(!ok)  puts("impossible");
        }
      }
      return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/7742801.html
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