HDU 6026 Deleting Edges【最短路】【思维题】

Deleting Edges

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 567 Accepted Submission(s): 210

Problem Description

Little Q is crazy about graph theory, and now he creates a game about graphs and trees.
There is a bi-directional graph with

n nodes, labeled from 0 to

n−1. Every edge has its length, which is a positive integer ranged from 1 to 9.
Now, Little Q wants to delete some edges (or delete nothing) in the graph to get a new graph, which satisfies the following requirements:
(1) The new graph is a tree with

n−1 edges.
(2) For every vertice

v(0<v<n), the distance between 0 and

v on the tree is equal to the length of shortest path from 0 to

v in the original graph.
Little Q wonders the number of ways to delete edges to get such a satisfied graph. If there exists an edge between two nodes

i and

j, while in another graph there isn't such edge, then we regard the two graphs different.
Since the answer may be very large, please print the answer modulo

109+7.

Input

The input contains several test cases, no more than 10 test cases.
In each test case, the first line contains an integer

n(1≤n≤50), denoting the number of nodes in the graph.
In the following

n lines, every line contains a string with

n characters. These strings describes the adjacency matrix of the graph. Suppose the

j-th number of the

i-th line is

c(0≤c≤9), if

c is a positive integer, there is an edge between

i and

j with length of

c, if

c=0, then there isn't any edge between

i and

j.
The input data ensure that the

i-th number of the

i-th line is always 0, and the

j-th number of the

i-th line is always equal to the

i-th number of the

j-th line.

Output

For each test case, print a single line containing a single integer, denoting the answer modulo

109+7.

Sample Input

Sample Output

1
6

Source

2017中国大学生程序设计竞赛 - 女生专场

题意：给你一个图，让你删掉一些边后变成一棵树，这棵树的根是0号节点，要满足在树上从0到任意节点的最短距离和原图相等。问总共有多少种删法？

做法：

用dijkstra求出原图中0点到每个点的最短路的长度，再暴力的跑一遍判断能否通过别的点同样使得0点到当前点的距离仍是最短路径，如果有一个可替代点，则表示对于当前点存在一种可以删边的方法，将所有的可能删的方法相乘即可。

#include<iostream>
#include<algorithm>
#include<cmath>
#include<queue>
#include<functional>
#define INF 0x3f3f3f3f
#define ms(a,b) memset(a,b,sizeof(a))

using namespace std;

typedef long long ll;

const int mod = 1e9 + 7;
const int maxn = 55;

int a[maxn][maxn];
int d[maxn], n;

void dij()
{
	ms(d, INF);
	d[0] = 0;
	bool book[maxn] = { 0 };
	while (1)
	{
		int v = -1;
		for (int i = 0; i < n; i++)
		{
			if (!book[i] && (v == -1 || d[i] < d[v]))
				v = i;
		}
		if (v == -1) break;
		book[v] = 1;
		for (int i = 0; i < n; i++)
		{
			if (!book[i])
			{
				if (d[i] > d[v] + a[v][i])
				{
					d[i] = d[v] + a[v][i];
				}
			}
		}
	}
}

void solve()
{
	ll ans = 1;
	for (int i = 1; i < n; i++)
	{
		ll tmp = 0;
		for (int j = 0; j < n; j++)
		{
			if (d[i] == d[j] + a[j][i])
			{
				tmp++;
			}
		}
		ans = (ans*tmp) % mod;
	}
	printf("%lld
", ans);
}

int main()
{
	while (~scanf("%d", &n))
	{
		for (int i = 0; i < n; i++)
		{
			for (int j = 0; j < n; j++)
			{
				char p;
				scanf(" %c", &p);
				int q = p - '0';
				if (q == 0)
					q = INF;
				a[i][j]= q;
			}
		}
		dij();
		solve();
	}
}

Fighting~