HDU3631

1.题目描述：

Shortest Path

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5443    Accepted Submission(s): 1332

Problem Description

When YY was a boy and LMY was a girl, they trained for NOI (National Olympiad in Informatics) in GD team. One day, GD team’s coach, Prof. GUO asked them to solve the following shortest-path problem.
There is a weighted directed multigraph G. And there are following two operations for the weighted directed multigraph:
(1) Mark a vertex in the graph.
(2) Find the shortest-path between two vertices only through marked vertices.
For it was the first time that LMY faced such a problem, she was very nervous. At this moment, YY decided to help LMY to analyze the shortest-path problem. With the help of YY, LMY solved the problem at once, admiring YY very much. Since then, when LMY meets problems, she always calls YY to analyze the problems for her. Of course, YY is very glad to help LMY. Finally, it is known to us all, YY and LMY become programming lovers.
Could you also solve the shortest-path problem?

 

Input

The input consists of multiple test cases. For each test case, the first line contains three integers N, M and Q, where N is the number of vertices in the given graph, N≤300; M is the number of arcs, M≤100000; and Q is the number of operations, Q ≤100000. All vertices are number as 0, 1, 2, … , N - 1, respectively. Initially all vertices are unmarked. Each of the next M lines describes an arc by three integers (x, y, c): initial vertex (x), terminal vertex (y), and the weight of the arc (c). (c > 0) Then each of the next Q lines describes an operation, where operation “0 x” represents that vertex x is marked, and operation “1 x y” finds the length of shortest-path between x and y only through marked vertices. There is a blank line between two consecutive test cases.
End of input is indicated by a line containing N = M = Q = 0.

 

Output

Start each test case with "Case #:" on a single line, where # is the case number starting from 1.
For operation “0 x”, if vertex x has been marked, output “ERROR! At point x”.
For operation “1 x y”, if vertex x or vertex y isn’t marked, output “ERROR! At path x to y”; if y isn’t reachable from x through marked vertices, output “No such path”; otherwise output the length of the shortest-path. The format is showed as sample output.
There is a blank line between two consecutive test cases.

 

Sample Input

5 10 101 2 63350 4 57253 3 69634 0 81461 2 99621 0 19432 1 23924 2 1542 2 74221 3 98960 10 30 20 40 40 11 3 31 1 10 30 40 0 0

 

Sample Output

Case 1:ERROR! At point 4ERROR! At point 100ERROR! At point 3ERROR! At point 4

2.题意概述：

给你一个图，然后接下来先描述图的情况。之后会有查询操作和标记操作，查询的点如果没有被标记则输出ERROR，否则输出在当前已标记节点条件下两路径的最短路

3.解题思路：

最开始想法真开vis数组，每次更新都进行一次floyd硬怼，但是很害怕，毕竟这样是n^4

8688091

Time Limit Exceeded

  

G++

1917103:34:32转念一想。等等，这不就是floyd的操作吗？每标记一个点就去更新这个点对图的影响。最后成功AC，最后想总结的是，对算法必须要学会理解，这道题典型就是考察对floyd算法的理解，这样才能活学活用啊。

4.AC代码：

#include <bits/stdc++.h>#define INF 0x1f1f1f1f#define maxn 100100#define N 333#define eps 1e-6#define pi acos(-1.0)#define e exp(1.0)using namespace std;const int mod = 1e9 + 7;typedef long long ll;typedef unsigned long long ull;bool vis[N];int mp[N][N];void floyd(int k, int n){    for (int i = 0; i < n; i++)        for (int j = 0; j < n; j++)            mp[i][j] = min(mp[i][j], mp[i][k] + mp[k][j]);}int main(){#ifndef ONLINE_JUDGE    freopen("in.txt", "r", stdin);    freopen("out.txt", "w", stdout);    long _begin_time = clock();#endif    int n, m, q, kase = 0;    while (~scanf("%d%d%d", &n, &m, &q), n + m + q)    {        if (kase)            puts("");        printf("Case %d:\n", ++kase);        for (int i = 0; i < n; i++)            for (int j = 0; j < n; j++)                mp[i][j] = i == j ? 0 : INF;        memset(vis, 0, sizeof(vis));        while (m--)        {            int u, v, w;            scanf("%d%d%d", &u, &v, &w);            mp[u][v] = min(mp[u][v], w);        }        while (q--)        {            int dir, u, v;            scanf("%d%d", &dir, &u);            if (dir == 0)            {                if (vis[u])                    printf("ERROR! At point %d\n", u);                else                {                    vis[u] = 1;                    floyd(u, n);                }            }            else            {                scanf("%d", &v);                if (vis[u] && vis[v])                {                    if (mp[u][v] == INF)                        puts("No such path");                    else                        printf("%d\n", mp[u][v]);                }                else                    printf("ERROR! At path %d to %d\n", u, v);            }        }    }#ifndef ONLINE_JUDGE    long _end_time = clock();    printf("time = %ld ms.", _end_time - _begin_time);#endif    return 0;}