HDU2833-最短路+dp

绝地灬酷狼 2022-07-11 13:10 150阅读 0赞

# WuKong #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)  
Total Submission(s): 1804    Accepted Submission(s): 672**

Problem Description

Liyuan wanted to rewrite the famous book “Journey to the West” (“Xi You Ji” in Chinese pinyin). In the original book, the Monkey King Sun Wukong was trapped by the Buddha for 500 years, then he was rescued by Tang Monk, and began his journey to the west. Liyuan thought it is too brutal for the monkey, so he changed the story:  
  
One day, Wukong left his home - Mountain of Flower and Fruit, to the Dragon   King’s party, at the same time, Tang Monk left Baima Temple to the Lingyin Temple to deliver a lecture. They are both busy, so they will choose the shortest path. However, there may be several different shortest paths between two places. Now the Buddha wants them to encounter on the road. To increase the possibility of their meeting, the Buddha wants to arrange the two routes to make their common places as many as possible. Of course, the two routines should still be the shortest paths.  
  
Unfortunately, the Buddha is not good at algorithm, so he ask you for help.

Input

There are several test cases in the input. The first line of each case contains the number of places N (1 <= N <= 300) and the number of roads M (1 <= M <= N\*N), separated by a space. Then M lines follow, each of which contains three integers a b c, indicating there is a road between place a and b, whose length is c. Please note the roads are undirected. The last line contains four integers A B C D, separated by spaces, indicating the start and end points of Wukong, and the start and end points of Tang Monk respectively.   
  
The input are ended with N=M=0, which should not be processed.

Output

Output one line for each case, indicating the maximum common points of the two shortest paths.

Sample Input

6 6
    1 2 1
    2 3 1
    3 4 1
    4 5 1
    1 5 2
    4 6 3
    1 6 2 4
    0 0

Sample Output

3
    
    Hint: One possible arrangement is (1-2-3-4-6) for Wukong and (2-3-4) for Tang Monk. The number of common points are 3.

**题目大意：**

给你一张无向图和两条路径的起点和终点，求都以最短路走时的最长公共点个数

**题目思路：**

我们很容易想到的是，因为都是走最短路，所以公共的点一定是连续的，既然是连续的

我们不妨用dp\[i\]\[j\]来表示从i到j的最短路经过的点有多少个，而处理这个我们可以用floyd

来处理这个，我们知道floyd跟新k点的时候是mp\[i\]\[j\]=mp\[i\]\[k\]+mp\[k\]\[j\],所以就相当于把ij拆

开接到k上，所以我们很好想到dp\[i\]\[j\]=dp\[i\]\[k\]+dp\[k\]\[j\]-1，因为k点重复了一次所以要减去1

处理多条路径时我们只需dp\[i\]\[j\]= max(dp\[i\]\[j\],dp\[i\]\[k\]+dp\[k\]\[j\]-1); 所以我们只需枚举所有的ij

的组合，而要使i->j是属于两条路径的我们可以想到当mp\[s\]\[i\]+mp\[i\]\[j\]+mp\[j\]\[e\]==mp\[s\]\[e\]时

可知ij是属于se的，因为这都是最短路径

**AC代码：**

#include<algorithm>
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    
    const int maxn = 5e2+10;
    
    const int inf = 1e9;
    
    int mp[maxn][maxn],dp[maxn][maxn];
    int n,m,s0,e0,s1,e1;
    
    void floyd(){
        for(int k=1;k<=n;k++)
            for(int i=1;i<=n;i++)
                for(int j=1;j<=n;j++){
                    if(mp[i][j]>mp[i][k]+mp[k][j]){
                        mp[i][j]=mp[i][k]+mp[k][j];
                        dp[i][j]=dp[i][k]+dp[k][j]-1;
                    }else if(mp[i][j]==mp[i][k]+mp[k][j]&&dp[i][j]<dp[i][k]+dp[k][j]){
                        dp[i][j]=dp[i][k]+dp[k][j]-1;
                    }
                }
    }
    
    void sove(){
        int ans = 0;
        for(int i=1;i<=n;i++)
            for(int j=1;j<=n;j++){
                if(mp[s0][i]+mp[i][j]+mp[j][e0]==mp[s0][e0]&&mp[s1][i]+mp[i][j]+mp[j][e1]==mp[s1][e1])
                    ans=max(dp[i][j],ans);
            }
        printf("%d\n",ans);
    }
    
    int main()
    {
        while(scanf("%d%d",&n,&m),n+m){
            for(int i=1;i<=n;i++)
                for(int j=1;j<=n;j++)
                    if(i!=j)mp[i][j]=inf,dp[i][j]=2;
                    else mp[i][j]=0,dp[j][j]=1;
            while(m--){
                int u,v,w;
                scanf("%d%d%d",&u,&v,&w);
                mp[u][v]=mp[v][u]=min(mp[u][v],w);
            }
            scanf("%d%d%d%d",&s0,&e0,&s1,&e1);
            floyd();
            sove();
        }
    
        return 0;
    }