[HDU6331]Problem M. Walking Plan

最新推荐文章于 2020-03-08 19:50:52 发布

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最新推荐文章于 2020-03-08 19:50:52 发布

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分类专栏： -----------------图论最短路 floyd ---------------------分治递推

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本文介绍了一种针对特定场景下的最短路径问题的优化算法，该算法通过分块技术预处理图中的路径信息，实现对多组询问的有效响应。特别适用于需要多次查询不同起点和终点的最短路径的情况。

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Time Limit: 5000/2500 MS (Java/Others)
Memory Limit: 524288/524288 K (Java/Others)

Problem Description
There are n intersections in Bytetown, connected with m one way streets. Little Q likes sport walking very much, he plans to walk for $q$ days. On the i-th day, Little Q plans to start walking at the $s_i-th$ intersection, walk through at least $k_i$ streets and finally return to the $t_i-th$ intersection.
Little Q’s smart phone will record his walking route. Compared to stay healthy, Little Q cares the statistics more. So he wants to minimize the total walking length of each day. Please write a program to help him find the best route.

Input
The first line of the input contains an integer $T(1≤T≤10)$ , denoting the number of test cases.
In each test case, there are 2 integers $n,m(2≤n≤50,1≤m≤10000)$ in the first line, denoting the number of intersections and one way streets.
In the next m lines, each line contains 3 integers $u_i,v_i,w_i(1≤u_i,v_i≤n,u_i≠v_i,1≤w_i≤10000)$ , denoting a one way street from the intersection $u_i$ to $v_i$ , and the length of it is $w_i$ .
Then in the next line, there is an integer $q(1≤q≤100000)$ , denoting the number of days.
In the next $q$ lines, each line contains 3 integers $s_i,t_i,k_i(1≤s_i,t_i≤n,1≤k_i≤10000)$ , describing the walking plan.

Output
For each walking plan, print a single line containing an integer, denoting the minimum total walking length. If there is no solution, please print $-1$ .

Sample Input

Sample Output

Source
2018 Multi-University Training Contest 3

题意：
一张有向有边权图，q组询问，每组询问a,b,k，询问从a到b至少经过k条边的最短路。

题解：
考虑分块，
$f[k][i][j]$ 表示刚好走 $k(k<=100)$ 步从 $i$ 走到 $j$ 的最短路程
$g[k][i][j]$ 表示刚好走 $100*k(k<=100)$ 步从 $i$ 走到 $j$ 的最短路程
$ro[k][i][j]$ 表示至少走 $k(k<=100)$ 步从 $i$ 走到 $j$ 的最短路程
$val[i][j]$ 表示从 $i$ 到 $j$ 直接连边的权值
$mp[i][j]$ 表示从 $i$ 到 $j$ 没有限制步数的最短路距离
可得：
$f[k][i][j]=min(f[k][i][j],f[k-1][i][mid]+val[mid][j])$
$g[k][i][j]=min(g[k][i][j],g[k-1][i][mid]+f[100][mid][j])$
$ro[k][i][j]=min(ro[k][i][j],f[k][i][mid]+mp[mid][j])$

答案为 $min(g[k/100][a][i]+ro[k$ $MOD$ $100][i][b])$ $i ∈[1,n]$

#include<bits/stdc++.h>
#define LiangJiaJun main
#define ll long long
#define INF 1050000000
using namespace std;
int val[54][54];
int f[104][54][54],g[104][54][54],ro[104][54][54],mp[54][54];
int n,m;
int pre(){
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            mp[i][j]=val[i][j];
        }
    }
    for(int i=1;i<=n;i++)mp[i][i]=0;
    for(int k=1;k<=n;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                mp[i][j]=min(mp[i][j],mp[i][k]+mp[k][j]);
            }
        }
    }

    for(int k=0;k<=100;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                f[k][i][j]=INF;
            }
        }
    }
    for(int i=1;i<=n;i++)f[0][i][i]=0;
    for(int k=1;k<=100;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                for(int mid=1;mid<=n;mid++){
                    f[k][i][j]=min(f[k][i][j],f[k-1][i][mid]+val[mid][j]);
                }
            }
        }

    }
    for(int k=0;k<=100;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                g[k][i][j]=INF;
            }
        }
    }
    for(int i=1;i<=n;i++)g[0][i][i]=0;
    for(int k=1;k<=100;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                for(int mid=1;mid<=n;mid++){
                    g[k][i][j]=min(g[k][i][j],g[k-1][i][mid]+f[100][mid][j]);
                }
            }
        }
    }
    for(int k=0;k<=100;k++){
        for(int i=1;i<=n;i++){
            for(int j=1;j<=n;j++){
                ro[k][i][j]=INF;
                for(int mid=1;mid<=n;mid++){
                    ro[k][i][j]=min(ro[k][i][j],f[k][i][mid]+mp[mid][j]);
                }
            }
        }
    }
    return 0;
}
int w33ha(){
    scanf("%d%d",&n,&m);
    for(int i=0;i<=51;i++)
        for(int j=0;j<=51;j++)
            val[i][j]=INF;
    for(int i=1;i<=m;i++){
        int u,v,w;
        scanf("%d%d%d",&u,&v,&w);
        val[u][v]=min(val[u][v],w);
    }
    pre();
    int q;scanf("%d",&q);
    while(q--){
        int a,b,k;
        scanf("%d%d%d",&a,&b,&k);
        int ans=INF;
        for(int mid=1;mid<=n;mid++){
            ans=min(ans,g[k/100][a][mid]+ro[k%100][mid][b]);
        }
        printf("%d\n",(ans==INF?-1:ans));
    }
    return 0;
}
int LiangJiaJun(){
    int T;scanf("%d",&T);
    while(T--)w33ha();
    return 0;
}