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，终于过了，这样的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 堆中的元素增加和减少值的操作

2 取出堆顶值的操作

灵活改动Dijsktra，只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的操作全部用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 = http://www.mamicode.com/msi[str];>



SPOJ 15. The Shortest Path 堆优化Dijsktra