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

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：

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 1000000007
using 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();
        //最后一个点就可以跳出了
        if(x.x==n-1)
            break;
        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
",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 1000000007

struct 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
",i,dis[i]);
//        }
        printf("%d
",ans);
        }
        return 0;
    }