You are given a list of cities. Each direct connection between two cities has its transportation cost (an integer bigger than 0). The goal is to find the paths of minimum cost between pairs of cities. Assume that the cost of each path (which is the sum of costs
of all direct connections belongning to this path) is at most 200000. The name of a city is a string containing characters a,...,z and is at most 10 characters long.

Input

s [the number of tests <= 10]
n [the number of cities <= 10000]
NAME [city name]
p [the number of neighbours of city NAME]
nr cost [nr - index of a city connected to NAME (the index of the first city is 1)]
[cost - the transportation cost]
r [the number of paths to find <= 100]
NAME1 NAME2 [NAME1 - source, NAME2 - destination]
[empty line separating the tests]

Output

cost [the minimum transportation cost from city NAME1 to city NAME2 (one per line)]

Example

Input:
1
4
gdansk
2
2 1
3 3
bydgoszcz
3
1 1
3 1
4 4
torun
3
1 3
2 1
4 1
warszawa
2
2 4
3 1
2
gdansk warszawa
bydgoszcz warszawa Output:
3
2

使用堆优化Dijsktra的代码都是一大坨的。写起来好累。

要求对堆和图论和Dijsktra算法都十分熟悉。

这次写了两个多小时,最终过了,这种题目对思维锻炼是十分有帮助的。

优先熟悉堆的主要函数有:

1 堆中的元素添加和降低值的操作

2 取出堆顶值的操作

灵活修改Dijsktra。仅仅是求两点之间的最短路径。

之前使用指针写过。这次使用静态数组和vector来表示邻接表来解决。不用指针动态分配内存,速度更加快点。

Heap的操作所实用class封装起来了。

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>
#include <string>
#include <map>
#include <vector>
#include <string.h> using namespace std;
const int MAX_C = 15;
const int MAX_N = 10005; struct Node
{
int des, cost;
};
vector<Node> gra[MAX_N]; void insertNeighbor(int src, int des, int cost)
{
Node n;
n.cost = cost;
n.des = des;
gra[src].push_back(n);
} struct hNode
{
int ver, dis;
};
hNode heaps[MAX_N];
int hPos[MAX_N];//指示顶点在堆中的位置 class MinHeap
{
public:
int size;
MinHeap(int s = 0): size(s) {} int lson(int rt) { return rt<<1; }
int rson(int rt) { return rt<<1 | 1; }
int parent(int rt) { return rt>>1; } void swaphNode(int l, int r)
{
hNode t = heaps[l];
heaps[l] = heaps[r];
heaps[r] = t; hPos[heaps[r].ver] = r;
hPos[heaps[l].ver] = l;
} void pushUp(int rt)
{
while (parent(rt) > 0 && heaps[parent(rt)].dis > heaps[rt].dis)
{
swaphNode(rt, parent(rt));
rt = parent(rt);
}
} void pushDown(int rt)
{
int l = lson(rt);
if (l > size) return ;
int r = rson(rt); int sma = rt;
if (heaps[sma].dis > heaps[l].dis) sma = l;
if (r <= size && heaps[sma].dis > heaps[r].dis) sma = r; if (sma != rt)
{
swaphNode(sma, rt);
pushDown(sma);
}
} void increase(int ver, int dis)
{
int rt = hPos[ver];
heaps[rt].dis = dis; pushDown(rt);
} void decrease(int ver, int dis)
{
int rt = hPos[ver];
heaps[rt].dis = dis; pushUp(rt);
} void insert(int ver, int dis)
{
size++;
heaps[size].dis = dis;
heaps[size].ver = ver;
hPos[ver] = size; pushUp(size);
} bool verIsInHeap(int ver)
{
int rt = hPos[ver];
return rt <= size;
} bool isInHeap(int rt)
{
return rt <= size;
} void extractMin()
{
swaphNode(1, size);
--size;
pushDown(1);
}
}; int dijsktra(int src, int des, int vers)
{
MinHeap mheap;
for (int v = 1; v <= vers; v++)
{
mheap.insert(v, INT_MAX);
}
mheap.decrease(src, 0); for (int v = 1; v < vers; v++)
{
if (heaps[1].ver == des) return heaps[1].dis;
int u = heaps[1].ver;
int dis = heaps[1].dis; if (dis == INT_MAX) return INT_MAX;//防止溢出 mheap.extractMin(); int n = (int)gra[u].size();
for (int j = 0; j < n; j++)
{
int ver = gra[u][j].des;
int c = gra[u][j].cost;
int rt = hPos[ver]; if (mheap.isInHeap(rt) && dis+c < heaps[rt].dis)
{
mheap.decrease(ver, dis+c);
}
}
}
return heaps[1].dis;
} int main()
{
int T, n, p, nr, cost, r, src, des;
scanf("%d", &T);
while (T--)
{
scanf("%d", &n);
memset(heaps, 0, sizeof(hNode) * (n+1));
memset(hPos, 0, sizeof(int) * (n+1));
for (int i = 0; i <= n; i++)
{
gra[i].clear();
} map<string, int> msi;
char str[MAX_C];
for (int i = 1; i <= n; i++)
{
scanf("%s", str);
msi[str] = i;
scanf("%d", &p);
for (int j = 0; j < p; j++)
{
scanf("%d %d", &nr, &cost);
insertNeighbor(i, nr, cost);
}
}
scanf("%d", &r);
for (int i = 0; i < r; i++)
{
scanf("%s", str);
src = msi[str];
scanf("%s", str);
des = msi[str]; printf("%d\n", dijsktra(src, des, n));
}
}
return 0;
}

版权声明:笔者靖心脏,景空间地址:http://blog.csdn.net/kenden23/,只有经过作者同意转载。

SPOJ 15. The Shortest Path 堆优化Dijsktra的更多相关文章

  1. SPOJ 15. The Shortest Path 最短路径题解

    本题就是给出一组cities.然后以下会询问,两个cities之间的最短路径. 属于反复询问的问题,临时我仅仅想到使用Dijsktra+heap实现了. 由于本题反复查询次数也不多,故此假设保存全部最 ...

  2. [CF1051F]The Shortest Statement_堆优化dij_最短路树_倍增lca

    The Shortest Statement 题目链接:https://codeforces.com/contest/1051/problem/F 数据范围:略. 题解: 关于这个题,有一个重要的性质 ...

  3. POJ-2387.Til the Cows Come Home.(五种方法:Dijkstra + Dijkstra堆优化 + Bellman-Ford + SPFA + Floyd-Warshall)

    昨天刚学习完最短路的算法,今天开始练题发现我是真的菜呀,居然能忘记邻接表是怎么写的,真的是菜的真实...... 为了弥补自己的菜,我决定这道题我就要用五种办法写出,并在Dijkstra算法堆优化中另外 ...

  4. [CF843D]Dynamic Shortest Path

    [CF843D]Dynamic Shortest Path 题目大意: 给定一个带权有向图,包含\(n(n\le10^5)\)个点和\(m(m\le10^5)\)条边.共\(q(q\le2000)\) ...

  5. NEU 1685: All Pair Shortest Path

    题目描述 Bobo has a directed graph G with n vertex labeled by 1,2,3,..n. Let D(i,j) be the number of edg ...

  6. PAT-1030 Travel Plan (30 分) 最短路最小边权 堆优化dijkstra+DFS

    PAT 1030 最短路最小边权 堆优化dijkstra+DFS 1030 Travel Plan (30 分) A traveler's map gives the distances betwee ...

  7. Codeforces Round #303 (Div. 2) E. Paths and Trees Dijkstra堆优化+贪心(!!!)

    E. Paths and Trees time limit per test 3 seconds memory limit per test 256 megabytes input standard ...

  8. 深入理解dijkstra+堆优化

    深入理解dijkstra+堆优化 其实就这几种代码几种结构,记住了完全就可以举一反三,所以多记多练多优化多思考. Dijkstra   对于一个有向图或无向图,所有边权为正(边用邻接矩阵的形式给出), ...

  9. hdu-----(2807)The Shortest Path(矩阵+Floyd)

    The Shortest Path Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others ...

随机推荐

  1. MAC地址格式小结

    之前一段时间在做网卡驱动的工作,如今产品量产,利用ifconfig eth hw ether在配置mac地址时发现一个问题, 随机配置一个mac地址,发现有的会报出Cannot assign requ ...

  2. OpenCV两张图片的合并

    转载请注明出处..! http://blog.csdn.net/zhonghuan1992 OpenCV两张图片的合并 原理: 两张图片合并,想想图片是用一个个像素点来存储.每一个像素点有他的值. 那 ...

  3. phpcms v9框架的目录结构分析

    phpcms v9框架的目录结构分析:      了解v9框架的目录结构,有助于帮助我们快速建立起对v9框架的一个整体认识 打开"mycms"项目,有如下文件和目录      使用 ...

  4. 黄聪:Microsoft Enterprise Library 5.0 系列教程(三) Validation Application Block (高级)

    原文:黄聪:Microsoft Enterprise Library 5.0 系列教程(三) Validation Application Block (高级) 企业库验证应用程序模块之配置文件模式: ...

  5. 第三章 AOP 基于Schema的AOP

    基于Schema定义的切面和前现两种方式定义的切面,内容上都差不多,只是表现形式不一样而已. 3.7.1一般增强的使用 a.目标类 public class Target { public void ...

  6. Android 调试native的crash和anr

    1. 于trace找到相应的库.例如 liba.so和相应的地址信息 2. 采用addr2line 查看 addr2line 住址 -e liba.so -f 要么 arm-eabi-addr2lin ...

  7. jquey :eq(1)

    $("#div_Goods .datagrid-row .numberbox:eq(1)") $("#div_Goods .datagrid-row .numberbox ...

  8. 【牛刀小试2】password保

    ]password保 主要知识: 1.        while循环 2.        do-while循环 3.        if-else 4.        strcmp()函数 [充电一下 ...

  9. springmvc如何访问静态文件,例如jpg,js,css

    你怎么DispatcherServlet拦截"*.do"这有一个后缀URL.就不存在訪问不到静态资源的问题.   假设你的DispatcherServlet拦截"/&qu ...

  10. SlopOne推荐算法

    在开源框架taste中有SlopOne的Java实现,效果不错.使用movielens的数据,代码例如以下 代码 #coding:utf-8 import re import math #读取数据,并 ...