2017 CCPC-WFinal&&HDOJ6026 Deleting Edges

Deleting Edges

TimeLimit: 2000/1000 MS (Java/Others)    Memory Limit:131072/131072 K (Java/Others)
Total Submission(s): 570    Accepted Submission(s): 212

Problem Description

LittleQ is crazy about graph theory, and now he creates a game about graphs andtrees.
There is a bi-directional graph with n nodes, labeledfrom 0 ton−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 toget 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 tov 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 nodesi andj, 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 withn characters. These strings describes the adjacency matrix of the graph. Suppose thej -th number of thei -th line isc(0≤c≤9), ifc is a positive  integer, there is an edge between i and j with length ofc, ifc=0, then there isn't any edge between i and  j.
The input data ensure that the i -th number of thei -th line is always 0, and the j -th number of the i -th line is always equal to thei -th number of thej -th line.

 

 

Output

For each test case, print a single line containing a single integer, denoting theanswer modulo 109+7.

 

 

Sample Input

2

01

10

4

0123

1012

2101

3210

 

 

Sample Output

1

6

题意：给出n个点以及它们的连接情况及边的权值，要求删除一些边后(可以不删除)，使剩下的点构成一颗生成树(只有n-1条边，且所有点两两可达)，且0到其余点的距离等于原来的图中它们之间的距离。

思路：我们先用dis[i]来表示0到点i的最短距离，由于要求删边前后距离不变，容易想到对于(i,j)两点，如果dis[i]+dis_(i->j)==dis[j]那么i到j的这一条边便可以作为我们要求的生成树的边。那么问题就转化为对应于每个点，有多少满足要求的边链接到它，有几条边就有几种情况，最后把每个点的情况数相乘即可。

上面说到的dis[i]只要跑一遍SPFA即可。

PS：在做求最短路类似的问题时，利用vector向量容器存边能显著提高运行效率

#include <cstdio>#include <vector>#include <queue>#include <cstring>#include <algorithm>using namespace std;#define mst(a,b) memset((a),(b),sizeof(a))#define f(i,a,b) for(int i=(a);i<=(b);++i)#define rush() int T;scanf("%d",&T);while(T--)typedef long long ll;const int maxn= 1005;const int mod = 1e9+7;const int INF = 0x3f3f3f3f;const double eps = 1e-6;char a[maxn][maxn];vector<int>vec[maxn];int vis[maxn];int dis[maxn];int num[maxn];void spfa(){    mst(vis,0);    mst(dis,0x3f);    dis[0]=0;    vis[0]=1;    queue<int>q;    q.push(0);    while(q.size())    {        int cur=q.front();        q.pop();        vis[cur]=0;        for(int i=0;i<vec[cur].size();i++)        {            int nex=vec[cur][i];            int w=a[cur][nex]-'0';            if(dis[nex]>dis[cur]+w)            {                dis[nex]=dis[cur]+w;                if(vis[nex]==0)                {                    vis[nex]=1;                    q.push(nex);                }            }        }    }}int main(){    int n;    while(~scanf("%d",&n))    {        mst(num,0);        for(int i=0;i<n;i++)        {            scanf("%s",a[i]);            vec[i].clear();        }        for(int i=0;i<n;i++)        {            for(int j=0;j<i;j++)            {                if(a[i][j]!='0')                {                    vec[i].push_back(j);                    vec[j].push_back(i);                }            }        }        spfa();        for(int i=0;i<n;i++)        for(int j=0;j<n;j++)        {            if(a[i][j]=='0')    continue;            int w=a[i][j]-'0';            if(dis[i]+w==dis[j])                num[j]++;        }        ll ans=1;        for(int i=1;i<n;i++)        {            ans=(ans*num[i])%mod;        }        printf("%I64d\n",ans);    }    return 0;}