UVA 10806 Dijkstra, Dijkstra.

 

大意：给你一个无向图，让你求到达终点的最短距离然后再返回求最短距离，但是不能经过同一条边两次。

思路：我的思路是去的时候求一次最短路，回来是将所有边反向，然后求一次最短路，测试数据过了，如果是有向图的话，应该可以AC，无向图不知为啥WA，欢迎大牛们指点。

等我学了最小费用流的时候再来看看。12.10.25

CODE：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
using namespace std;

#define MAXN 1000001
#define INF 0x3f3f3f3f

struct Edge
{
    int u, v, next;
    int w;
}edge[4*MAXN], edge2[4*MAXN];

int first[MAXN], first2[MAXN];
int d[MAXN];
int n, m, cnt;

void init()
{
    cnt = 0;
    memset(first, -1, sizeof(first));
    memset(first2, -1, sizeof(first));
}

void spfa(int src)
{
    queue<int> q;
    bool inq[MAXN] = {0};
    for(int i = 1; i <= n; i++) d[i] = (i == src)? 0:INF;
    q.push(src);
    while(!q.empty())
    {
        int x = q.front(); q.pop();
        inq[x] = 0;
        for(int e = first[x]; e != -1; e = edge[e].next)
        {
            int v = edge[e].v, w = edge[e].w;
            if(d[v] > w + d[x])
            {
                d[v] = d[x] + w;
                if(!inq[v])
                {
                    inq[v] = 1;
                    q.push(v);
                }
            }
        }
    }
}

void spfa2(int src)
{
    queue<int> q;
    bool inq[MAXN] = {0};
    for(int i = 1; i <= n; i++) d[i] = (i == src)? 0:INF;
    q.push(src);
    while(!q.empty())
    {
        int x = q.front(); q.pop();
        inq[x] = 0;
        for(int e = first2[x]; e != -1; e = edge2[e].next)
        {
            int v = edge2[e].v, w = edge2[e].w;
            if(d[v] > d[x] + w)
            {
                d[v] = d[x] + w;
                if(!inq[v])
                {
                    inq[v] = 1;
                    q.push(v);
                }
            }
        }
    }
}

void read_graph(int u, int v, int w) //正反向图 
{
    edge[cnt].v = v, edge[cnt].w = w;
    edge[cnt].next = first[u], first[u] = cnt;
    edge2[cnt].v = u, edge2[cnt].w = w;
    edge2[cnt].next = first2[v], first2[v] = cnt++;
}

void solve()
{
    int ans = 0;
    spfa(1);
    ans += d[n];
    if(ans >= INF)
    {
        printf("Back to jail\n");
        return ;
    }
    spfa2(1);
    ans += d[n];
    if(ans >= INF)
    {
        printf("Back to jail\n");
    }
    else
    {
        printf("%d\n", ans);
    }
}

int main()
{
    while(scanf("%d%d", &n, &m))
    {
        init();
        while(m--)
        {
            int u, v, w;
            scanf("%d%d%d", &u, &v, &w);
            read_graph(u, v, w);
        }
        solve();
    }
    return 0;
}