HDU 5636 Shortest Path

Problem Description

There is a path graph  G=(V,E)  with  n  vertices. Vertices are numbered from  1  to  n  and there is an edge with unit length between  i  and  i+1   (1≤i<n) . To make the graph more interesting, someone adds three more edges to the graph. The length of each new edge is  1 .

You are given the graph and several queries about the shortest path between some pairs of vertices.

 

Input

There are multiple test cases. The first line of input contains an integer  T , indicating the number of test cases. For each test case:

The first line contains two integer  n  and  m   (1≤n,m≤105)  -- the number of vertices and the number of queries. The next line contains 6 integers  a1,b1,a2,b2,a3,b3   (1≤a1,a2,a3,b1,b2,b3≤n) , separated by a space, denoting the new added three edges are  (a1,b1) ,  (a2,b2) ,  (a3,b3) .

In the next  m  lines, each contains two integers  si  and  ti   (1≤si,ti≤n) , denoting a query.

The sum of values of  m  in all test cases doesn't exceed  106 .

 

Output

For each test cases, output an integer  S=(∑i=1mi⋅zi) mod (109+7) , where  zi  is the answer for  i -th query.

 

Sample Input

1 10 2 2 4 5 7 8 10 1 5 3 1

 

Sample Output

7

题解说的是用floyd，我是直接dfs，常数上差了点，原来在bc里面还能过的，现在外面过不了了。

#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<iostream>
#include<queue>
#include<map>
#include<algorithm>
using namespace std;
typedef long long LL;
const int INF = 0x7FFFFFFF;
const int mod = 1e9 + 7;
const int maxn = 1e5 + 10;
int T, n, m, s, t, ans, res;
int a[6], f[6];

void dfs(int s, int dis)
{
    ans = min(ans, dis + abs(s - t));
    for (int i = 0; i < 6; i++)
    if (f[i])
    {
        f[i] = f[i ^ 1] = 0;
        dfs(a[i ^ 1], dis + abs(s - a[i]) + 1);
        f[i] = f[i ^ 1] = 1;
    }
}

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d", &n, &m);
        for (int i = 0; i < 6; i++) scanf("%d", &a[i]), f[i] = 1;
        res = 0;
        for (int i = 1; i <= m; i++)
        {
            scanf("%d%d", &s, &t);
            ans = INF;
            dfs(s, 0);
            (res += (LL)ans*i%mod) %= mod;
        }
        printf("%d\n", res);
    }
    return 0;
}

现在补上floyd的，这题数据不多，用读入优化的意义不大

#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long LL;
const int maxn=1e5+10;
const int mod=1e9+7;
int T,n,m,a[6],dis[6][6],x,y;

int main()
{
	scanf("%d",&T);
	while (T--)
	{
		scanf("%d%d",&n,&m);
		for (int i=0;i<6;i++) scanf("%d",&a[i]);
		for (int i=0;i<6;i++) 
			for (int j=0;j<6;j++) dis[i][j]=abs(a[i]-a[j]);
		for (int i=0;i<6;i+=2) dis[i][i^1]=dis[i^1][i]=1;
		for (int i=0;i<6;i++)
			for (int j=0;j<6;j++)
				for (int k=0;k<6;k++)
					dis[j][k]=min(dis[j][k],dis[j][i]+dis[i][k]);
		int res=0;
		for (int k=1;k<=m;k++)
		{
			scanf("%d%d",&x,&y);
			int ans=abs(x-y);
			for (int i=0;i<6;i++)
				for (int j=0;j<6;j++)
				{
					ans=min(ans,abs(x-a[i])+dis[i][j]+abs(y-a[j]));
				}
			(res+=(LL)k*ans%mod)%=mod;
		}
		printf("%d\n",res);
	}
	return 0;
}