HDU 3986 Harry Potter and the Final Battle

Problem Description

The final battle is coming. Now Harry Potter is located at city 1, and Voldemort is located at city n. To make the world peace as soon as possible, Of course, Harry Potter will choose the shortest road between city 1 and city n. But unfortunately, Voldemort is so powerful that he can choose to destroy any one of the existing roads as he wish, but he can only destroy one. Now given the roads between cities, you are to give the shortest time that Harry Potter can reach city n and begin the battle in the worst case.

Input

First line, case number t (t<=20). 
Then for each case: an integer n (2<=n<=1000) means the number of city in the magical world, the cities are numbered from 1 to n. Then an integer m means the roads in the magical world, m (0< m <=50000). Following m lines, each line with three integer u, v, w (u != v,1 <=u, v<=n, 1<=w <1000), separated by a single space. It means there is a bidirectional road between u and v with the cost of time w. There may be multiple roads between two cities.

Output

Each case per line: the shortest time to reach city n in the worst case. If it is impossible to reach city n in the worst case, output “-1”.

Sample Input

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

Sample Output

15
-1
2
先求出最短路，然后从终点返回删边再算一遍最短路，记下最大值。
#include<stdio.h>
#include<string.h>
#include<vector>
#include<queue>
#include<functional>
using namespace std;
const int maxn = 10005;
int n, m, x, y, z, dis[maxn], v[5 * maxn], Dis[maxn], ans;

struct abc
{
    int next, v;
    abc(){};
    abc(int a, int b) :next(a), v(b){};
};

vector<abc> tree[maxn];

struct cmp
{
    bool operator ()(const int &a, const int &b)
    {
        return dis[a] > dis[b];
    }
};

int dijkstra(int begin, int end, int ex)
{
    priority_queue<int, vector<int>, cmp> p;
    memset(dis, -1, sizeof(dis));
    dis[begin] = 0;
    p.push(begin);
    while (!p.empty())
    {
        int q = p.top();    p.pop();
        if (q == end) break;
        for (int i = 0, j; i < tree[q].size(); i++)
        {
            j = tree[q][i].next;
            if (tree[q][i].v != ex)
            if (dis[j] == -1 || dis[j]> dis[q] + v[tree[q][i].v])
            {
                dis[j] = dis[q] + v[tree[q][i].v];
                p.push(j);
            }
        }
    }
    return dis[end];
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d", &n, &m);
        for (int i = 0; i <= n; i++) tree[i].clear();
        for (int i = 0; i < m; i++)
        {
            scanf("%d%d%d", &x, &y, &v[i]);
            tree[x].push_back(abc(y, i));
            tree[y].push_back(abc(x, i));
        }
        ans = dijkstra(1, n, -1);
        memcpy(Dis, dis, sizeof(dis));
        for (int i = n; i != 1;)
        {
            for (int j = 0; j < tree[i].size(); j++)
            {
                if (Dis[i] == Dis[tree[i][j].next] + v[tree[i][j].v])
                {
                    int k = dijkstra(1, n, tree[i][j].v);
                    if (k == -1) ans = -1; else ans = max(ans, k);
                    i = tree[i][j].next;
                    break;
                }
            }
            if (ans < 0) break;
        }
        printf("%d\n", ans);
    }
    return 0;
}