zoukankan      html  css  js  c++  java
  • AHU-412 Bessie Come Home 【Dijkstra】

    Description
    It's dinner time, and the cows are out in their separate pastures. Farmer John rings the bell so they will start walking to the barn. Your job is to figure out which one cow gets to the barn first (the supplied test data will always have exactly one fastest cow).

    Between milkings, each cow is located in her own pasture, though some pastures have no cows in them. Each pasture is connected by a path to one or more other pastures (potentially including itself). Sometimes, two (potentially self-same) pastures are connected by more than one path. One or more of the pastures has a path to the barn. Thus, all cows have a path to the barn and they always know the shortest path. Of course, cows can go either direction on a path and they all walk at the same speed.

    The pastures are labeled `a'..`z' and `A'..`Y'. One cow is in each pasture labeled with a capital letter. No cow is in a pasture labeled with a lower case letter. The barn's label is `Z'; no cows are in the barn, though.
    Input
    Line 1: Integer P (1 <= P <= 10000) the number of paths that interconnect the pastures (and the barn)
    Line 2..P+1: Space separated, two letters and an integer: the names of the interconnected pastures/barn and the distance between them (1 <= distance <= 1000)
    Output
    A single line containing two items: the capital letter name of the pasture of the cow that arrives first back at the barn, the length of the path followed by that cow.
     
    Sample Input
    5
    A d 6
    B d 3
    C e 9
    d Z 8
    e Z 3
    Sample Output
    B 11
    
    

    单源最短路+Dijkstra,细节处理要注意。


    Code:
     1 #include<bits/stdc++.h>
     2 using namespace std;
     3 #define M(a, b) memset(a, b, sizeof(a))
     4 #define INF 0x3f3f3f3f
     5 const int maxn = 100;
     6 int d[maxn];
     7 bool done[maxn];
     8 bool cow[maxn];
     9 struct Edge {
    10     int from, to, dist;
    11 };
    12 struct Node {
    13     int d, u;
    14     bool operator < (const Node& rhs) const {
    15         return d > rhs.d;
    16     }
    17 };
    18 vector<int> G[maxn];
    19 vector<Edge> edges;
    20 
    21 void init() {
    22     memset(cow, 0, sizeof(cow));
    23     for (int i = 0; i <= 52; ++i) G[i].clear();
    24     edges.clear();
    25 }
    26 
    27 void AddEdge(int from, int to, int dist) {
    28     edges.push_back(Edge{from, to, dist});
    29     int sz = edges.size();
    30     G[from].push_back(sz-1);
    31 }
    32 
    33 void Dijkstra(int s) {
    34     memset(done, 0, sizeof(done));
    35     memset(d, INF, sizeof(d));
    36     priority_queue<Node> Q;
    37     d[s] = 0;
    38     Q.push(Node{0, s});
    39     while(!Q.empty()) {
    40         Node x = Q.top(); Q.pop();
    41         int u = x.u;
    42         if (done[u]) continue;
    43         done[u] = true;
    44         int len = G[u].size();
    45         for (int i = 0; i < len; ++i) {
    46             Edge& e = edges[G[u][i]];
    47             if (d[e.to] > d[u] + e.dist) {
    48                 d[e.to] = d[u] + e.dist;
    49                 Q.push(Node{d[e.to], e.to});
    50             }
    51         }
    52     }
    53 } 
    54 
    55 int main() {
    56     int n, aa, bb, dd;
    57     char a[3], b[3];
    58     while(~scanf("%d", &n)) {
    59         init();
    60         for (int i = 0; i < n; ++i) {
    61             scanf("%s%s%d", a, b, &dd);
    62             if (a[0] <= 'Z') aa = a[0] - 'A' + 26, cow[aa] = 1;
    63             else aa = a[0] - 'a';
    64             if (b[0] <= 'Z') bb = b[0] - 'A' + 26, cow[bb] = 1;
    65             else bb = b[0] - 'a';
    66             AddEdge(aa, bb, dd);
    67             AddEdge(bb, aa, dd);
    68         }
    69         Dijkstra(51);
    70         int ans, ansd = INF;
    71         for (int i = 26; i < 51; ++i)
    72             if (cow[i] && d[i] < ansd) {
    73                 ansd = d[i];
    74                 ans = i;
    75             }
    76         printf("%c %d
    ", ans-26+'A', ansd);
    77     }
    78 
    79     return 0;
    80 }
  • 相关阅读:
    ABAP 销售范围
    C语言深度剖析自测题8解析
    Ubutun13.10下安装fcitx
    各种Web服务器与Nginx的对比
    Hadoop2.2.0在Ubuntu编译失败解决方法
    网站上在给出邮箱地址时不直接使用@符号的原因
    IPython notebook在浏览器中显示不正常的问题及解决方法
    input只读效果
    阿里实习生电面题目:输出给定字符串的全部连续子串
    DNS 放大
  • 原文地址:https://www.cnblogs.com/robin1998/p/6481508.html
Copyright © 2011-2022 走看看