nyoj 1002 Trucking

同样一道改编题。

只要把题意理解了好。

简单的二分加最短路。

只要二分高度，

然后求最短路，输出满足题意的即可。

代码如下：

（最短路用spfa 时间效率高）

 
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define inf 0x3fffffff
#define M 1005

struct edge
{
    int v, w, h, next;
} e[2000005];

int pre[M], cnt, dist[M], n;
bool inq[M];

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

void addedge (int u, int v, int w, int h)    //慢慢模拟就会明白的
{
    e[cnt].v = v;
    e[cnt].w = w;
    e[cnt].h = h;
    e[cnt].next = pre[u];    //接替已有边
    pre[u] = cnt++;          //自己成为u派生的第一条边
}

void spfa (int u, int lim)
{
    int v, w, i;
    for (i = 1; i <= n; i++)
        dist[i] = inf, inq[i] = false;
    dist[u] = 0;
    queue q;
    q.push (u);
    inq[u] = true;
    while (!q.empty())
    {
        u = q.front();
        q.pop();
        inq[u] = false;
        for (i = pre[u]; i != -1; i = e[i].next)
        {
            if (e[i].h < lim) continue;
            w = e[i].w;
            v = e[i].v;
            if (dist[u] + w < dist[v])
            {
                dist[v] = dist[u] + w;
                if (!inq[v])
                {
                    q.push (v);
                    inq[v] = true;
                }
            }
        }
    }
}

int main()
{
    //freopen("in2.txt","r",stdin);
    //freopen("in1.txt","w",stdout);
    int m, u, v, w, h, l, r, mid, cc = 1, res;
    while (scanf ("%d%d", &n, &m), (n||m))
    {
        if (cc > 1) printf ("\n");
        init();
        while (m--)
        {
            scanf ("%d%d%d%d", &u, &v, &h, &w);
            if (h == -1) h = inf;
            addedge (u, v, w, h);
            addedge (v, u, w, h);    //双向前插加边
        }
        scanf ("%d%d%d", &u, &v, &h);
        l = 0, r = h, res = inf;
        while (l < r)    //简单二分枚举高度，使高度尽量大
        {
            mid = (l+r+1) >> 1;
            spfa (u, mid);
            if (dist[v] != inf)
                l = mid, res = dist[v];
            else r = mid - 1;
        }
        printf ("Case %d:\n", cc++);
        if (res != inf)
            printf ("maximum height = %d\nlength of shortest route = %d\n", l, res);
        else puts ("cannot reach destination");
    }
    //printf("TIME used = %.4lf\n",(double)clock()/CLOCKS_PER_SEC);
    return 0;
}