CodeForces 601A （dijkstra算法）

Destroying Roads
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d & %I64u
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Status

Practice

CodeForces 543B
Description
In some country there are exactly n cities and m bidirectional roads connecting the cities. Cities are numbered with integers from 1 to n. If cities a and b are connected by a road, then in an hour you can go along this road either from city a to city b, or from city b to city a. The road network is such that from any city you can get to any other one by moving along the roads.

You want to destroy the largest possible number of roads in the country so that the remaining roads would allow you to get from city s1 to city t1 in at most l1 hours and get from city s2 to city t2 in at most l2 hours.

Determine what maximum number of roads you need to destroy in order to meet the condition of your plan. If it is impossible to reach the desired result, print -1.

Input
The first line contains two integers n, m (1 ≤ n ≤ 3000, ) — the number of cities and roads in the country, respectively.

Next m lines contain the descriptions of the roads as pairs of integers ai, bi (1 ≤ ai, bi ≤ n, ai ≠ bi). It is guaranteed that the roads that are given in the description can transport you from any city to any other one. It is guaranteed that each pair of cities has at most one road between them.

The last two lines contains three integers each, s1, t1, l1 and s2, t2, l2, respectively (1 ≤ si, ti ≤ n, 0 ≤ li ≤ n).

Output
Print a single number — the answer to the problem. If the it is impossible to meet the conditions, print -1.

Sample Input
Input
5 4
1 2
2 3
3 4
4 5
1 3 2
3 5 2
Output
0
Input
5 4
1 2
2 3
3 4
4 5
1 3 2
2 4 2
Output
1
Input
5 4
1 2
2 3
3 4
4 5
1 3 2
3 5 1
Output
-1

题意：给你 n 个点，m 条火车路线，该路径所用都是 1，剩下的多少汽车路线，让你求它们分别从 1 走，问它们都到达 n 点的最短时间。
题解：定义两个数组，一个存火车的路径，一个存汽车的路径，然后求最短路径。再比较一下哪个用时长就取哪个；

#include 
#include 
#include 
using namespace std;
#define M 500
#define INF 0x3f3f3f3f

int map_t[M][M], map_b[M][M], n, m;
int dist[M];
int dijkstra(int map_s[][M])
{
    bool vis[M];
    memset(vis, 0, sizeof(vis));
    for(int i=0; i<=n; i++) dist[i] = INF;
    for(int i=1; i<=n; i++)
    {
        dist[i] = map_s[1][i];
    }
    vis[1] = 1;
    for(int i=1; i<=n; i++)
    {
        int mins = INF, _next = 1;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j] && mins > dist[j])
            {
                _next = j;
                mins = dist[j];
            }
        }
        vis[_next] = 1;
        for(int j=1; j<=n; j++)
        {
            if(!vis[j] && dist[j] > mins + map_s[_next][j])
            {
                dist[j] = mins + map_s[_next][j];
            }
        }
    }
    return dist[n];
}
void init()//初始化
{
    for(int i=0; i<=n; i++)
    {
        for(int j=0; j<=n; j++)
        {
            if(i == j)
            {
                map_t[i][j] = 0;
                map_b[i][j] = 0;
            }
            else
            {
                map_t[i][j] = INF;
                map_b[i][j] = 1;
            }
        }
    }
}
int main()
{
    while(scanf("%d%d", &n, &m) != EOF)
    {
        init();
        int a, b;
        for(int i=0; iscanf("%d%d", &a, &b);
            map_t[a][b] = map_t[b][a] = 1;
            map_b[a][b] = map_b[b][a] = INF;
        }
        int anst = dijkstra(map_t);
        int ansb = dijkstra(map_b);
        if(anst == INF || ansb == INF)
        {
            printf("-1\n");
        }
        else
        {
            ansb = max(anst, ansb);
            printf("%d\n", ansb);
        }
    }

    return 0;
}