Deleting Edges
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 0 Accepted Submission(s): 0
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
题意：有n个点，告诉你两两之间的边，让你去掉一些边，使得0号节点到其他节点都是最短路，问有多少种这样的图
解题思路：最短路，先建一个只包含最短路的有向无环图，每一个点选择任意一条入边即可生成一个树形图，那么树的种类就等于每个点的入度乘积
#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>
using namespace std;
#define LL long long
const int INF=0x3f3f3f3f;
#define mod 1000000007
int n,s[60],nt[5009],e[5009],l[5009],visit[60],dis[60];
LL a[60];
char ch[60][60];
struct node
{
int id,l;
friend bool operator <(node a,node b)
{
return a.l>b.l;
}
};
void Dijkstra()
{
memset(visit,0,sizeof visit);
memset(dis,INF,sizeof dis);
priority_queue<node>q;
node pre,nt1;
pre.id=0,pre.l=0,a[0]=1;
dis[0]=0;
q.push(pre);
while(!q.empty())
{
pre=q.top();
q.pop();
visit[pre.id]=1;
for(int i=s[pre.id];~i;i=nt[i])
{
int ee=e[i];
if(visit[ee]) continue;
if(dis[ee]>pre.l+l[i])
{
a[ee]=1;
dis[ee]=pre.l+l[i];
nt1.id=ee;
nt1.l=dis[ee];
q.push(nt1);
}
else if(dis[ee]==pre.l+l[i])
a[ee]++;
}
}
LL ans=1;
for(int i=0;i<n;i++)
{
ans*=a[i];
ans%=mod;
}
printf("%lld\n",ans);
}
int main()
{
while(~scanf("%d",&n))
{
int cnt=0;
memset(s,-1,sizeof s);
for(int i=0;i<n;i++)
{
scanf("%s",ch[i]);
for(int j=0;j<n;j++)
{
if(ch[i][j]!='0')
nt[cnt]=s[i],s[i]=cnt,e[cnt]=j,l[cnt++]=ch[i][j]-'0';
}
}
Dijkstra();
}
return 0;
}
