HDU 6026 Deleting Edges（读懂题意+spfa）

Deleting Edges

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1325    Accepted Submission(s): 455

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

2 01 10 4 0123 1012 2101 3210

 

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，包含n个点若干条带权边，你要随便挑n-1条构成树，且挑出来的边在原图中是到根节点0的距离最小的边。

读题是关键。

先spfa预处理出最短的边。

然后记录每个节点包含的最短的边的数目。

最后乘起来就行了，相当于每个点选一条最短的边，一共就是n-1条。注意long long和mod。

AC代码：

#include <bits/stdc++.h>
#define ll long long
using namespace std;
int const MAX = 200005;
int const INF = 1 << 30;
const ll mo=1e9+7;
int n, m;

struct EDGE
{
    int u, v;
    int val;
}e[2000010];
struct NODE
{
    int v, w;
    NODE(int vv, int ww)
    {
        v = vv;
        w = ww;
    }
};

vector <NODE> vt[MAX];
int dis[MAX];
bool vis[MAX];

void SPFA(int v0)
{
    memset(vis, false, sizeof(vis));
    for(int i = 1; i <= n; i++)
        dis[i] = INF;
    dis[v0] = 0;
    queue <int> q;
    q.push(v0);
    while(!q.empty())
    {
        int u = q.front();
        q.pop();
        vis[u] = false;
        int sz = vt[u].size();
        for(int i = 0; i < sz; i++)
        {
            int v = vt[u][i].v;
            int w = vt[u][i].w;
            if(dis[v] > dis[u] + w)
            {
                dis[v] = dis[u] + w;
                if(!vis[v])
                {
                    q.push(v);
                    vis[v] = true;
                }
            }
        }
    }
}
char s[60][60];
ll ans[10010];
int main()
{
    while(scanf("%d",&n)!=EOF)
    {
        for(int i=0;i<=n;i++)
        vt[i].clear();
        for(int i=0;i<n;i++) scanf("%s",s[i]);
        for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
        {
            int d=(s[i][j]&15);
            if(d)
            {
                vt[i].push_back(NODE(j, d));
            }
        }
        SPFA(0);
        memset(ans,0,sizeof(ans));
        for(int u=0;u<n;u++)
        {
        int sz = vt[u].size();
        for(int i = 0; i < sz; i++)
        {
            int v = vt[u][i].v;
            int w = vt[u][i].w;
            if(dis[v]==dis[u] + w)
            {
                ans[v]++;
            }
        }
        }
        ll anss=1;
        for(int i=1;i<n;i++)
        {
            //cout<<ans[i]<<endl;
            anss=(anss*ans[i])%mo;
        }
        printf("%lld\n",anss);
    }
    return 0;
}