Luo Gu P1613 foot

Algorithms DPTags: 图论, ,倍增

topic

Topic background

A little work is not only cumbersome, more stringent requirement that every morning A small company arrives before 6:00, otherwise cleared salary this month. A small but happens to have a bad habit of stay in bed. So in order to keep his salary, bought a small A foot space is a very cow B, every second can run \ (2 ^ k \) one thousand meters ( \ (k \) is any natural number). Of course, this machine longint is stored, so the total foot length can not exceed maxlongint km. A small home to the company can be viewed as a way to view a small point A home, a company point n, the length of each edge are one thousand meters. A small day want to be able to wake up as late as possible, so let him help you calculate, he needs at least a few seconds to get to the company. A data path to ensure that at least 1 to n.

Title Description

The first line of two integers n, m, the number indicates the number of points and edges.

The next two lines each m numbers u, v, v represents a u to the edge.

Input Format

Line a number that represents the minimum number of seconds to the company.

Output Format

Conveying the least amount of output.

Sample input and output

Input # 1

Output # 1

Description / Tips

Sample [explain]

1-> 1-> 2-> 3-> 4, the total path length of 4 km and a foot control can be used directly.

【data range】

50% of the data satisfies the optimal solution path length <= 1000;

100% data satisfies n <= 50, m <= 10000, optimal solution path length <= maxlongint.

answer:

Starting from the title, when you see the title of " \ (2 ^ k \) " when you can think of this question to a skill - doubled .

Suppose we have some point, if the distance between two points where the sum may reach \ (n-^ \ 2) , then the distance between these two points is the cost \ (1 \) , so that the entire figure, All distance \ (2 ^ n \) edge weight of the edge between the pair of points is 1. The question then becomes how to ask the right side which between two points is 1 .

Here we use the DP to achieve multiplication, we use the outermost loop k=1->64, represents power, the inner loop 2 i=1->n, , j=1->n, t=1->nwhether or not a point can be updated by an edge weight of 1 reaches the edge, thereby realizing multiplication process , specific code as follows:

for (int k = 1; k <= 64; k ++ )
    for (int i = 1; i <= n; i ++ )
        for (int j = 1; j <= n; j ++ )
            for (int t = 1; t <= n; t ++ )
                if (f[j][i][k - 1] && f[i][t][k - 1])
                    f[j][t][k] = dis[j][t] = 1;

After the data satisfying the title \ (n-\ Le 50 \) , so ultimately the shortest path can be used when the Floyd , three Solving the shortest cycle, the shortest path to the final output between from 1-n.

AC Code

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
using namespace std;
const int N = 70;
int n, m;
int f[N][N][N], dis[N][N];
int main()
{
    memset(dis, 0x3f, sizeof dis);
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= m; i ++ )
    {
        int x, y;
        scanf("%d%d", &x, &y);
        f[x][y][0] = 1;
        dis[x][y] = 1;
    }
    for (int k = 1; k <= 64; k ++ )
        for (int i = 1; i <= n; i ++ )
            for (int j = 1; j <= n; j ++ )
                for (int t = 1; t <= n; t ++ )
                    if (f[j][i][k - 1] && f[i][t][k - 1])
                        f[j][t][k] = dis[j][t] = 1;
    for (int k = 1; k <= n; k ++ )
        for (int i = 1; i <= n; i ++)
            for (int j = 1; j <= n; j ++ )
                dis[i][j] = min(dis[i][k] + dis[k][j], dis[i][j]);
    printf("%d", dis[1][n]);
    return 0;
}

Luo Gu P1613 foot - DP, graph theory