HDU Problem 5636 Shortest Path 【Floyd】

Shortest Path

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1587    Accepted Submission(s): 513

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

 

Source

BestCoder Round #74 (div.2)

 

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题意：有一个节点数为n的链表，每个节点相距一个单位的长度，但是有三个传送门，问你每次给出两点的最短距离是多少

思路：如果没有传送门的存在，距离是abs（x - y），但是现在多了传送门，总共也就6个，可以预处理每个传送门的距离，两次for循环，试一下每次从传送门a进入，再从任意一个出去的距离。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
#define space " "
using namespace std;
//typedef long long LL;
typedef __int64 Int;
typedef pair<int,int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double Pi = acos(-1);
const int MOD = 1e9 + 7;
const int MAXN = 1e5 + 10;
Int d[7][7], a[7];
Int abs_long(Int x) {
    if (x < 0) return -x;
    return x;
}
int main() {
    int t, m, n;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &m);
        for (int i = 0; i < 3; i++) {
            scanf("%I64d%I64d", &a[i], &a[i + 3]);
        }
        //初始化传送门之间的距离
        for (int i = 0; i < 6; i++) {
            for (int j = 0; j < 6; j++) {
                d[i][j] = abs_long(a[i] - a[j]);
            }
        }
        for (int i = 0; i < 3; i++) {
            d[i][i + 3] = d[i + 3][i] = 1;
        }
        for (int k = 0; k < 6; k++) {
            for (int i = 0; i < 6; i++) {
                for (int j = 0; j < 6; j++) {
                    d[i][j] = min(d[i][j], d[i][k] + d[k][j]);
                }
            }
        }
        Int x, y; Int ans = 0, temp, sum;
        for (int  k = 1; k <= m; k++) {
            scanf("%I64d%I64d", &x, &y);
            temp = abs_long(x - y);
            //用每一个传送门检查最短距离
            for (int i = 0; i < 6; i++) {
                for (int j = 0; j < 6; j++) {
                    sum = abs_long(x - a[i]) + d[i][j] + abs_long(y - a[j]);
                    temp = min(sum, temp);
                }
            }
            ans = (ans + temp*k)%MOD;
        }
        printf("%I64d
", ans);
    }
    return 0;
}