UVA 10269 Adventure of Super Mario

UVA_10269

    由于马里奥的飞行距离有限，因此为了方便处理，我们首先用floyd预处理出马里奥可以飞行的两点间的最短路，然后再将图分成K+1层用SPFA求最短路即可。

#include<stdio.h>
#include<string.h>
#define MAXD 130
#define MAXN 20
#define INF 1000000000
int A, B, M, L, K, N;
int G[MAXD][MAXD], d[MAXD][MAXN], st[MAXD * MAXN];
int q[MAXD * MAXN], inq[MAXD][MAXN];
void init()
{
    int i, j, k, a, b, t;
    scanf("%d%d%d%d%d", &A, &B, &M, &L, &K);
    N = A + B;
    for(i = 1; i <= N; i ++)
        for(j = 1; j <= N; j ++)
        {
            if(i == j)
                G[i][j] = 0;
            else
                G[i][j] = INF;
        }
    for(i = 0; i < M; i ++)
    {
        scanf("%d%d%d", &a, &b, &t);
        G[a][b] = G[b][a] = t;
    }
}
void floyd()
{
    int i, j, k;
    for(k = 1; k <= N; k ++)
        for(i = 1; i <= N; i ++)
            for(j = 1; j <= N; j ++)
                if(k <= A && G[i][k] + G[k][j] < G[i][j])
                    G[i][j] = G[i][k] + G[k][j];
}
int SPFA()
{
    int i, j, k, u, v, front, rear, max = (K + 1) * N;
    front = rear = 0;
    memset(inq, 0, sizeof(0));
    for(i = 1; i <= N; i ++)
        for(j = 0; j <= K; j ++)
            d[i][j] = INF;
    d[N][0] = 0;
    q[rear] = N;
    st[rear] = 0;
    rear ++;
    while(front != rear)
    {
        u = q[front];
        k = st[front];
        front ++;
        if(front > max)
            front = 0;
        inq[u][k] = 0;
        for(v = 1; v <= N; v ++)
        {
            if(G[u][v] + d[u][k] < d[v][k])
            {
                d[v][k] = G[u][v] + d[u][k];
                if(!inq[v][k])
                {
                    q[rear] = v;
                    st[rear] = k;
                    rear ++;
                    if(rear > max)
                        rear = 0;
                    inq[v][k] = 1;
                }
            }
            if(G[u][v] <= L && k < K && d[u][k] < d[v][k + 1])
            {
                d[v][k + 1] = d[u][k];
                if(!inq[v][k + 1])
                {
                    q[rear] = v;
                    st[rear] = k + 1;
                    rear ++;
                    if(rear > max)
                        rear = 0;
                    inq[v][k + 1] = 1;
                }
            }
        }
    }
    int res = INF;
    for(i = 0; i <= K; i ++)
        if(d[1][i] < res)
            res = d[1][i];
    return res;
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        init();
        floyd();
        int res = SPFA();
        printf("%d\n", res);
    }
    return 0;
}