Deleting Edges
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)Total Submission(s): 1685 Accepted Submission(s): 572
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.
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.
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
做过两遍的原题比赛的时候愣是没写出来,真是日了狗了!!!
题意:给你一个含n个节点(n<=50,编号0~n-1)无向图,你要删去一些边,满足:
1、剩下的图是连通的。
2、剩下的图边数最少
3、原图中每个节点到源点0点的最短路上的边不能删。
求一共有多少种删法。
思路: 首先很容易想到,每一种删法对应原图的一颗生成树,因此我们只需要求有多少颗生成树即可。
于是先预处理出所有到源点0点的最短距离。
对于边(u,v),若dis[u]+len[u][v]==dis[v],则sum[v]++; 即从v点出发的可选边增加一条。
最后每个节点都选出一条出发的可选边即可组成一颗目标树,因此只需要乘起来累计答案即可。别忘了mod和long long
比赛的时候死活过不了,赛后补题1A,原因是赛场上写成了单向边。。。真tmd。。。
AC代码:
/* 我就是吃屎,也要A了这道题 */ #include<bits/stdc++.h> #define inf 0x3f3f3f3f #define ll long long using namespace std; const int maxn=110; const ll mo=1e9+7; int n,m; struct node { int to,de; }; ll sum[maxn]; vector<node>g[maxn]; bool book[maxn]; int dis[maxn],d[maxn][maxn]; void init() { memset(book,0,sizeof(book)); memset(sum,0,sizeof(sum)); for(int i=0;i<maxn;i++) g[i].clear(); memset(dis,inf,sizeof(dis)); } void spfa(int s) { queue<int>q; q.push(s); book[s]=1; while(!q.empty()) { int pt=q.front(); q.pop(); book[pt]=0; for(int i=0;i<g[pt].size();i++) { int tmp=g[pt][i].de+dis[pt]; if(tmp<dis[g[pt][i].to]) { dis[g[pt][i].to]=tmp; if(!book[g[pt][i].to]) { q.push(g[pt][i].to); book[g[pt][i].to]=1; } } } } } char s[maxn][maxn]; int main() { int i,x,y,j,c,T,cas; while(scanf("%d",&n)!=EOF) { init(); m=n; for(int i=0;i<m;i++) { scanf("%s",s[i]); } for(int i=0;i<n;i++) for(int j=0;j<n;j++) if(i!=j){ int x=(s[i][j]&15); if(!x) continue; node k; k.de=x; k.to=j; g[i].push_back(k); } dis[0]=0; spfa(0); for(int i=0;i<n;i++) for(int j=0;j<n;j++)//比赛的时候这里写成了i+1开始,无脑WA,真是日了狗了!!! if(j!=i){ int x=(s[i][j]&15); if(!x) continue; if(dis[i]+x==dis[j]) sum[j]++; } ll ans=1; for(int i=1;i<n;i++) ans=(ans*sum[i])%mo; printf("%lld\n",ans); } return 0; }
不要为了做题而做题,做过的题目一定要理解透彻!!!