WA UVa 10806 - Dijkstra, Dijkstra.

无向图，求从1出发到n再回到1的最短路，不能重复走同一条边。

错误思路：先求出1到n的最短路，删去经过的边，再求一次最短路，相加,容易找到反例。

WA

# include <cstdio>

# include <cstring>

# include <queue>

# include <algorithm>



using namespace std;



# define N 100 + 5

# define INF 0x3c3c3c3c



int n, m;

int pre[N], d[N];

int w[N][N];



void destory(void)

{

    for (int x = n; x != 1; x = pre[x])

    {

        w[x][pre[x]] = w[pre[x]][x] = INF;

    }

}



void read_graph(void)

{    

    int x, y, z;

    scanf("%d", &m);

    for (int i = 1; i <= n; ++i) 

        memset(w[i]+1, 0x3c, sizeof(int)*n);

    for (int i = 1; i <= m; ++i)

    {

        scanf("%d%d%d", &x, &y, &z);

        w[x][y] = w[y][x] = z;

    }

}



int dijkstra(int src, int des)

{

    typedef pair <int, int> pii;

    for (int i = 1; i <= n; ++i) d[i] = (i==src ? 0:INF);

    priority_queue < pii, vector<pii>, greater<pii> > Q;

    Q.push(make_pair(d[src], src));

    while (!Q.empty())

    {

        pii cur = Q.top(); Q.pop();

        int x = cur.second;

        if (x == des) return d[des];

        if (cur.first != d[x]) continue;

        for (int y = 1; y <= n; ++y) if (d[y] > d[x]+w[x][y])

        {

            d[y] = d[x] + w[x][y];

            pre[y] = x;

            Q.push(make_pair(d[y], y));

        }

    }

    return d[des];

}



int main()

{

    int ans;

        

    int icase = 0;

    while (scanf("%d", &n), n)

    {

        ++icase;

        if (icase != 1) putchar('\n');

        ans = 0;

        read_graph();

        ans += dijkstra(1, n);

        destory();

        ans += dijkstra(1, n);

        if (ans >= INF)

            printf("Back to jail");

        else

            printf("%d", ans);

    }



    return 0;

}

这是网络流的题目。