HDU 2017女生赛04 (变形最短路)

Deleting Edges

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 263    Accepted Submission(s): 85

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

2011040123101221013210

 

Sample Output

16

 

Source

2017中国大学生程序设计竞赛 - 女生专场

 

题意：

n个点，n*2-n条边，删除掉只剩下n-1条边，满住剩下的每个点在原图中都是最短路的存在。

分析：

dij队列优化需要vis，可以避免重复入队的情况，但是不加vis也不会无尽循环下去。

来自某个博客其他人的解法，其实也不需要跑完全图。

说入度乘积我更倾向于说成每个点走法的乘积。（5.15重温这句话发现这句话还是有问题的，到三点的还有一种画法）

dij算法就是求原点到各个点的最短路，也很贴合这道题的性质。那么不同走法之间会有影响吗？只有两个独立事件同时发生没有影响才能相乘。（有问题）

结果是没有影响。原题中说只要存在任意一条不同的边就是不同的图。

dij最短路的原理就是通过不同路来松弛。每种走法的组合一定可以存在一条异边。第三个点也是建立在前面点基础上通过扩展而来。

因为n为50，所以随便哪个最短路算法都能过吧：

正插邻接表加队列的确比反插好写多了。。。

#include <iostream>#include <cstdio>#include <queue>#include <vector>#include <bits/stdc++.h>#define mod 1000000007using namespace std;typedef long long ll;const int maxn = 100+10;const int INF = 0x3f3f3f3f;char ch[60][60];int cnt[maxn];struct node{    int x,d;    node(){}    node(int a,int b){x=a;d=b;}    bool operator < (const node & a) const    {        return d > a.d;    }};vector<node> eg[maxn];int dis[maxn];void Dijkstra(int s){    dis[s]=0;    //用优先队列优化    priority_queue<node> q;    q.push(node(s,dis[s]));    while(!q.empty())    {        node x=q.top();q.pop();        for(int i=0;i<eg[x.x].size();i++)        {            node y=eg[x.x][i];            if(dis[y.x]>x.d+y.d)            {                cnt[y.x] = 1;                dis[y.x]=x.d+y.d;                q.push(node(y.x,dis[y.x]));            }            else if(dis[y.x]==x.d+y.d) {                cnt[y.x]++;            }        }    }}int main(){//    freopen("in.txt","r",stdin);    int n;    while(~scanf("%d",&n))    {        for(int i=0;i<=n;i++)            dis[i]=INF;        memset(cnt,0,sizeof(cnt));        cnt[0]=1;        for(int i=0;i<=n;i++) eg[i].clear();            for(int i=0;i<n;i++)            {                scanf("%s",ch[i]);                for(int j=0;j<n;j++)                {                    if(ch[i][j]!='0')                        eg[i].push_back(node(j,ch[i][j]-'0'));                }            }        Dijkstra(0);        ll ans = 1LL;        for(int i = 0; i < n; i++) {            ans *= cnt[i];            ans %= mod;        }        printf("%d\n",ans);    }    return 0;}

一开始超时的，不知道在哪里：

反正以后就用vector写了。。。。

#include <cstdio>#include <cmath>#include <cctype>#include <algorithm>#include <cstring>#include <utility>#include <string>#include <iostream>#include <map>#include <set>#include <vector>#include <queue>#include <stack>typedef long long ll;using namespace std;const int maxn = 60+10;const int INF = 0x3f3f3f3f;//int mp[maxn][maxn];int first[maxn];int cnt[maxn];int num,dis[100];char ch[60][60];int vis[60];#define mod 1000000007struct Node {    int id;    int val;    }node;struct Edge {    int id;//以此点为出边找边    int val;    int next;    }e[maxn];void add(int u,int v,int d) {    //num边的编号    e[num].id  = v;    e[num].val = d;    e[num].next = first[u];    first[u] = num;    num++;}priority_queue<Node> q;bool operator < (Node a,Node b) {    return a.val > b.val;}int main() {//    freopen("in.txt","r",stdin);    int n,m;    while(~scanf("%d",&n)) {        memset(first,-1,sizeof(first));        memset(cnt,0,sizeof(cnt));        while(!q.empty()) q.pop();        for(int i=0;i<n;i++)        {            scanf("%s",ch[i]);            for(int j=0;j<n;j++)            {                if(ch[i][j]!='0')                    add(i,j,ch[i][j]-'0');            }        }            for(int i = 1; i <= n; i++) {                dis[i] = INF;            }            Node cur;            dis[0] = 0;            node.id = 0;            node.val = 0;            q.push(node);            cnt[0] = 1;            while(!q.empty()) {//                if(cur.id == End){//                    break;//            }                cur = q.top();                q.pop();                //i为边的编号                for(int i = first[cur.id]; i != -1; i = e[i].next) {                    if(dis[e[i].id] > e[i].val+cur.val) {                        cnt[e[i].id] = 1;                        dis[e[i].id] = e[i].val+cur.val;                        node.id = e[i].id;                        node.val = dis[e[i].id];                        q.push(node);                    }                    else if(dis[e[i].id] == e[i].val+cur.val)                        cnt[e[i].id]++;                }        }        ll ans = 1LL;        for(int i = 0; i < n; i++) {            ans *= cnt[i];            ans %= mod;        }//        for(int i = 0; i <= n; i++) {//            printf("初始点到%d点的距离为%d\n",i,dis[i]);//        }        printf("%d\n",ans);        }        return 0;    }