SPOJ 15. The Shortest Path 堆优化Dijsktra
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
- 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 ...
- [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 ...
- 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 ...
- [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\(q(q\le2000)\) ...
- 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 ...
- 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 ...
- 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 ...
- 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), ...
- 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
- CentOS7 编译安装LNMP
(文章来自:http://www.cnblogs.com/i-it/p/3841840.html,请各位到这个网址去看原文的) LNMP(Linux-Nginx-Mysql-PHP),本文在CentO ...
- 从零開始学android<ImageSwitcher图片切换组件.二十六.>
ImageSwitcher组件的主要功能是完毕图片的切换显示,比如用户在进行图片浏览的时候.能够通过button点击一张张的切换显示的图片,并且使用ImageSwitcher组件在每次切换的时候也能够 ...
- Android-Universal-Image-Loader学习笔记(一个)
Android-Universal-Image-Loader它是一个开源项目,处理图像加载和缓存.闲暇的时候,读一些源.特别记录. 所述图像文件(磁盘)高速缓存,我们需要考虑的因素,如以下 1) 定 ...
- windows phone (25) Canvas元素B
原文:windows phone (25) Canvas元素B ZIndex 这也是一个附加属性,表示canvas的children集合内的子元素的显示顺序,在canvas中的元素默认情况下是后面的 ...
- Prototype Pattern 原型模式
7.6 原型模式总结 原型模式作为一种快速创建大量相同或相似对象的方式,在软件开发中应用较为广泛,很多软件提供的复制(Ctrl + C)和粘贴(Ctrl + V)操作就是原型模式的典型应用,下面对该模 ...
- Java 并发专题 : CyclicBarrier 打造一个安全的门禁系统
继续并发专题~ 这次介绍CyclicBarrier:看一眼API的注释: /** * A synchronization aid that allows a set of threads to all ...
- 学派Delphi方法(推荐)——————————【Badboy】
Delphi是一个新的可视化编程环境, 提供了一种方便.快捷的Windows使用顺序开发工具. 它使用了MicrosoftWindows图形用户界面的很多先进特性和设计思想. 本文就给读者引见学Del ...
- 重新想象 Windows 8 Store Apps (18) - 绘图: Shape, Path, Stroke, Brush
原文:重新想象 Windows 8 Store Apps (18) - 绘图: Shape, Path, Stroke, Brush [源码下载] 重新想象 Windows 8 Store Apps ...
- 【原创】leetCodeOj --- Merge k Sorted Lists 解题报告
题目地址: https://oj.leetcode.com/problems/merge-k-sorted-lists/ 题目内容: /** * Definition for singly-linke ...
- 【原创】《算法导论》链表一章带星习题试解——附C语言实现
原题: 双向链表中,需要三个基本数据,一个携带具体数据,一个携带指向上一环节的prev指针,一个携带指向下一环节的next指针.请改写双向链表,仅用一个指针np实现双向链表的功能.定义np为next ...