PAT All Roads Lead to Rome 单源最短路
思路:单源最短路末班就好了,字符串映射成数字处理。
AC代码
//#define LOCAL
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <map>
#include <vector>
#include <string>
using namespace std;
#define inf 0x3f3f3f3f
const int maxn = 200;
map<string, int> ha;
int id, st, ed;
int n, k;
int hap[maxn];
char names[maxn][50];
int getId(string s) {
    if(!ha.count(s)) {
        ha[s] = id++;
    }
    return ha[s];
}
struct Edge {
    int from, to, dist;
    Edge(int u, int v, int d):from(u),to(v),dist(d) {}
};
vector<Edge> edges;
vector<int> G[maxn];
bool done[maxn];
int d[maxn], hp[maxn], pt[maxn], routes[maxn]; //Best
int p[maxn];
void init() {
    ha.clear();
    id = 0;
    for(int i = 0; i < maxn; i++) G[i].clear();
    edges.clear();
}
void addEdge(int from, int to, int dist) {
    edges.push_back(Edge(from, to, dist));
    int m = edges.size();
    G[from].push_back(m-1);
}
struct HeapNode{
    int d, u;
    HeapNode(int d, int u):d(d), u(u){
    }
    bool operator < (const HeapNode& rhs) const {
        return d > rhs.d;
    }
};
//距离小,开心多,平均大
void dijkstra(int s) {
    memset(done, 0, sizeof(done));
    priority_queue<HeapNode> Q;
    for(int i = 0; i < n; i++) {
        d[i] = inf;
        hp[i] = -inf;
        pt[i] = inf;
    }
    d[s] = hp[s] = pt[s] = 0;
    routes[s] = 1;
    Q.push(HeapNode(0, s));
    while(!Q.empty()) {
        HeapNode x = Q.top(); Q.pop();
        int u = x.u;
        if(done[u]) continue;
        done[u] = true;
        for(int i = 0; i < G[u].size(); i++) {
            Edge& e = edges[G[u][i]];
            bool update = false;
            if(d[e.to] > d[u] + e.dist) {
                routes[e.to] = routes[u];
                update = true;
            } else if(d[e.to] == d[u] + e.dist) {
                routes[e.to] += routes[u];
                if(hp[e.to] < hp[u] + hap[e.to]) {
                    update = true;
                } else if(hp[e.to] == hp[u] + hap[e.to]) {
                    if(pt[e.to] > pt[u] + 1) {
                        update = true;
                    }
                }
            }
            if(update) {
                d[e.to] = d[u] + e.dist;
                p[e.to] = u;
                hp[e.to] = hp[u] + hap[e.to];
                pt[e.to] = pt[u] + 1;
                Q.push(HeapNode(d[e.to], e.to));
            }
        }
    }
}
void print(int u) {
    if(u == 0) {
        printf("%s", names[0]);
        return;
    } else {
        print(p[u]);
        printf("->%s", names[u]);
    }
}
int main() {
#ifdef LOCAL
    freopen("data.in", "r", stdin);
    freopen("data.out", "w", stdout);
#endif
    while(scanf("%d%d%s", &n, &k, names[0]) == 3) {
        init();
        st = getId(names[0]);
        int happy;
        for(int i = 1; i < n; i++) {
            scanf("%s %d", names[i], &happy);
            int u = getId(names[i]);
            hap[u] = happy;
        }
        ed = getId("ROM");
        char x[50], y[50];
        int u, v, cost;
        for(int i = 0; i < k; i++) {
            scanf("%s%s%d", x, y, &cost);
            u = getId(x), v = getId(y);
            //printf("%d %d\n", u, v);
            addEdge(u, v, cost);
            addEdge(v, u, cost);
        }
        dijkstra(0);
        printf("%d %d %d %d\n", routes[ed], d[ed], hp[ed], (int)(hp[ed]/pt[ed]));
        print(ed);
        printf("\n");
    }
    return 0;
}
如有不当之处欢迎指出!
PAT All Roads Lead to Rome 单源最短路的更多相关文章
- PAT 1087 All Roads Lead to Rome[图论][迪杰斯特拉+dfs]
		
1087 All Roads Lead to Rome (30)(30 分) Indeed there are many different tourist routes from our city ...
 - PAT 1087 All Roads Lead to Rome
		
PAT 1087 All Roads Lead to Rome 题目: Indeed there are many different tourist routes from our city to ...
 - PAT甲级1087. All Roads Lead to Rome
		
PAT甲级1087. All Roads Lead to Rome 题意: 确实有从我们这个城市到罗马的不同的旅游线路.您应该以最低的成本找到您的客户的路线,同时获得最大的幸福. 输入规格: 每个输入 ...
 - PAT 甲级 1087  All Roads Lead to Rome(SPFA+DP)
		
题目链接 All Roads Lead to Rome 题目大意:求符合题意(三关键字)的最短路.并且算出路程最短的路径有几条. 思路:求最短路并不难,SPFA即可,关键是求总路程最短的路径条数. 我 ...
 - pat1087. All Roads Lead to Rome (30)
		
1087. All Roads Lead to Rome (30) 时间限制 200 ms 内存限制 65536 kB 代码长度限制 16000 B 判题程序 Standard 作者 CHEN, Yu ...
 - PAT_A1087#All Roads Lead to Rome
		
Source: PAT A1087 All Roads Lead to Rome (30 分) Description: Indeed there are many different tourist ...
 - [图的遍历&多标准] 1087. All Roads Lead to Rome (30)
		
1087. All Roads Lead to Rome (30) Indeed there are many different tourist routes from our city to Ro ...
 - PAT1087. All Roads Lead to Rome
		
PAT1087. All Roads Lead to Rome 题目大意 给定一个图的边权和点权, 求边权最小的路径; 若边权相同, 求点权最大; 若点权相同, 则求平均点权最大. 思路 先通过 Di ...
 - 最短路模板(Dijkstra & Dijkstra算法+堆优化 & bellman_ford  & 单源最短路SPFA)
		
关于几个的区别和联系:http://www.cnblogs.com/zswbky/p/5432353.html d.每组的第一行是三个整数T,S和D,表示有T条路,和草儿家相邻的城市的有S个(草儿家到 ...
 
随机推荐
- Overload&Override
			
Overload&Override overload-–重载 方法的重载就是在一个类中,可以定义多个有相同名字,但参数不同的方法.调用时,会根据不同的参数表选择对应的方法. 规 则:两同 ...
 - Log4j源码解析--框架流程+核心解析
			
OK,现在我们来研究Log4j的源码: 这篇博客有参照上善若水的博客,原文出处:http://www.blogjava.net/DLevin/archive/2012/06/28/381667.htm ...
 - CentOS如何把deb转为rpm
			
说明:可以转换,但不一定可用,可以根据报错提示,安装需要的依赖. 1 安装alien工具,下载地址http://ftp.de.debian.org/debian/pool/main/a/alien/ ...
 - Linux make nginx 的时候报错
			
报错如下: `conf/koi-win' and `/usr/local/nginx/conf/koi-win' are the same file 原因: 可能在编译 nginx 的时候步骤不对 ...
 - PhpStudy 升级 MySQL 版本到5.7
			
1:备份当前数据库数据. 最好是导成 SQL 文件 2:备份 PhpStudy 下的 MySQL 文件夹.以防升级失败.还可以使用旧版本的数据库 3:下载MySQL5.7.解压.然后放在 PhpStu ...
 - php与HTML交互问题
			
1.将表单中的action属性值设为PHP路径,则网页会跳转到这个网址 <html> <body> <form action="welcome.php" ...
 - httpd添加新模块
			
*/ .hljs { display: block; overflow-x: auto; padding: 0.5em; color: #333; background: #f8f8f8; } .hl ...
 - Div+Css画太极图源代码
			
<!DOCTYPE html><html> <head> <meta charset="utf-8"> <title>D ...
 - final修饰符,多态,抽象类,接口
			
1:final关键字(掌握) (1)是最终的意思,可以修饰类,方法,变量. (2)特点: A:它修饰的类,不能被继承. B:它修饰的方法,不能被重写. ...
 - Tomcat部署war应用总结
			
前言:罗列在Tomcat部署web应用的几种方法,供以后翻阅,其中的以helloapp为例 Tomcat目录介绍 简单目录介绍: bin目录:包含tomcat启动/关闭等脚本,支持linux.wind ...