POJ 1724 ROADS

POJ_1724

    这个题目可以直接将SPFA中的dis[]变成二维的dis[i][k]表示到第i个点花费为k时的最短路，但如果直接做SPFA会超时，需要加一个剪枝：设mind[i]表示到第i个点的用于剪枝最短路，mink[i] 表示第i个点的用于剪枝的最小花费，如果在SPFA的过程中走到了第i个点的路径长为d花费为k，那么如果d>=mind[i]&&k>=mink[i]，这时就进行剪枝，如果d<mind[i]&&k<mink[i]就更新mind[i]与mink[i]的值，并将这个点(i,k)入队，其余的情况也要将(i,k)入队。

    这个题目还可以用堆形式的SPFA去做，这时需要把N个点看成N*(K+1)个点，且堆顶的元素为当前所有点中dis[]的值最小的一个。其余的操作和SPFA大体相同，这时同样有一个剪枝：如果堆顶的元素是标号为N的点，那么就可以直接结束SPFA了。

//普通的SPFA
#include<stdio.h>
#include<string.h>
#define MAXD 110
#define MAXK 10010
#define MAXM 10010
#define MAXQ 1000110
#define INF 0x3f3f3f3f
int N, M, K, first[MAXD], next[MAXM], v[MAXM], w[MAXM], cost[MAXM], e, q[MAXQ], m[MAXQ], inq[MAXD][MAXK], d[MAXD][MAXK], mind[MAXD], mink[MAXD], Q = 1000100;
void add(int &x, int &y, int &z, int &t)
{
    v[e] = y;
    w[e] = z;
    cost[e] = t;
    next[e] = first[x];
    first[x] = e;
    ++ e;
}
void init()
{
    int i, j, k, x, y, z, t;
    e = 0;
    memset(first, -1, sizeof(first));
    scanf("%d%d", &N, &M);
    for(i = 0; i < M; i ++)
    {
        scanf("%d%d%d%d", &x, &y, &z, &t);
        add(x, y, z, t);
    }
}
void solve()
{
    int i, j, k, x, y, z, newk, front ,rear, ans;
    front = rear = 0;
    for(i = 1; i <= N; i ++)
        for(j = 0; j <= K; j ++)
            d[i][j] = INF, inq[i][j] = 0;
    memset(mind, 0x3f, sizeof(mind));
    memset(mink, 0x3f, sizeof(mink));
    q[rear] = 1, m[rear] = 0, d[1][0] = 0;
    ++ rear;
    mind[1] = mink[1] = 0;
    while(front != rear)
    {
        x = q[front], k = m[front];
        inq[x][k] = 0;
        ++ front;
        if(front > Q)
            front = 0;
        for(i = first[x]; i != -1; i = next[i])
        {
            y = v[i], newk = k + cost[i], z= d[x][k] + w[i];
            if(newk <= K && z < d[y][newk])
            {
                if(z >= mind[y] && newk >= mink[y])
                    continue;
                if(z < mind[y] && newk < mink[y])
                    mind[y] = z, mink[y] = newk;
                d[y][newk] = z;
                if(!inq[y][newk])
                {
                    q[rear] = y;
                    m[rear] = newk;
                    inq[y][newk] = 1;
                    ++ rear;
                    if(rear > Q)
                        rear = 0;
                }
            }
        }
    }
    ans = INF;
    for(i = 0; i <= K; i ++)
        if(d[N][i] < ans)
            ans = d[N][i];
    printf("%d\n", ans == INF ? -1 : ans);
}
int main()
{
    while(scanf("%d", &K) == 1)
    {
        init();
        solve();
    }
    return 0;
}

//堆形式的SPFA
#include<stdio.h>
#include<string.h>
#define MAXD 110
#define MAX 4000410
#define MAXM 10010
#define INF 0x3f3f3f3f
int N, M, K, first[MAXD], next[MAXM], v[MAXM], w[MAXM], cost[MAXM], P, f[MAX], d[MAX], m[MAX], tree[MAX], mind[MAXD], mink[MAXD];
void init()
{
    int i, j, k, x, y, z, t;
    memset(first, -1, sizeof(first));
    scanf("%d%d", &N, &M);
    for(i = 0; i < M; i ++)
    {
        scanf("%d%d%d%d", &x, &y, &z, &t);
        v[i] = y;
        w[i] = z;
        cost[i] = t;
        next[i] = first[x];
        first[x] = i;
    }
}
void insert(int cur, int dis, int k)
{
    int i, left, right, s = k * N + cur;
    if(dis < d[s])
    {
        f[s] = cur;
        d[s] = dis;
        m[s] = k;
        i = P + s;
        if(!tree[i])
        {
            tree[i] = s;
            for(i >>= 1; i > 0; i >>= 1)
            {
                left = tree[2 * i], right = tree[2 * i + 1];
                tree[i] = d[left] < d[right] ? left : right;
            }
        }
    }
}
void update()
{
    int i, left, right, s = tree[1];
    i = P + s;
    tree[i] = 0;
    for(i >>= 1; i > 0; i >>= 1)
    {
        left = tree[2 * i], right = tree[2 * i + 1];
        tree[i] = d[left] < d[right] ? left : right;
    }
}
void solve()
{
    int i, j, k, x, y, z, s, t, ans;
    memset(mind, 0x3f, sizeof(mind));
    memset(mink, 0x3f, sizeof(mink));
    for(P = 1; P < (K + 1) * N + 2; P <<= 1);
    for(i = (K + 1) * N; i >= 0; i --)
        d[i] = INF;
    for(i = 2 * P - 1; i > 0; i --)
        tree[i] = 0;
    d[0] = m[0] = INF;
    insert(1, 0, 0);
    mind[1] = mink[1] = 0;
    while(tree[1])
    {
        s = tree[1];
        if(s % N == 0)
            break;
        update();
        x = f[s], k = m[s];
        for(i = first[x]; i != -1; i = next[i])
        {
            y = v[i], z = d[s] + w[i], t = k + cost[i];
            if(t <= K)
            {
                if(z >= mind[y] && t >= mink[y])
                    continue;
                if(z < mind[y] && t < mink[y])
                    mind[y] = z, mink[y] = t;
                insert(y, z, t);
            }
        }
    }
    printf("%d\n", tree[1] ? d[s] : -1);
}
int main()
{
    while(scanf("%d", &K) == 1)
    {
        init();
        solve();
    }
    return 0;
}