House Man 【HDU - 3440】【差分约束】

题目链接

  就是我们必须跳N-1次，从最小的房子跳到最高的房子，然后呢，求最小的房子和最高的房子之间的最长的可能距离。那么就是差分约束咯。

  我们可以这么推，首先，对于所有的点，a[i] - a[i-1] >= 1，那么转换一下，就是a[i-1] - a[i] <= -1，然后看到之后的N-1步跳，每次都是跳到那个刚好比它大的那个房子上面，所以有a[x] - a[y] <= D，x、y分别代表了序号大的和序号小的，这两个之间的距离是"≤D"的。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define efs 1e-6
#define INF 0x3f3f3f3f
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MAX_3(a, b, c) max(max(a, b), c)
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN = 1e3 + 7, maxE = 1e4 + 7;
int N, D, cnt, head[maxN], dist[maxN], used[maxN];
struct node
{
    int val, id;
}a[maxN];
inline bool cmp(node e1, node e2) { return e1.val < e2.val; }
struct Eddge
{
    int nex, to, val;
    Eddge(int a=-1, int b=0, int c=0):nex(a), to(b), val(c) {}
}edge[maxE];
inline void addEddge(int u, int v, int w)
{
    edge[cnt] = Eddge(head[u], v, w);
    head[u] = cnt++;
}
queue<int> Q;
bool inque[maxN];
inline int spfa(int st, int ed)
{
    memset(inque, false, sizeof(inque));
    memset(dist, INF, sizeof(dist)); memset(used, 0, sizeof(used));
    dist[st] = 0;
    while(!Q.empty()) Q.pop();
    Q.push(st); inque[st] = true; used[st]++;
    while(!Q.empty())
    {
        int u = Q.front(); Q.pop(); inque[u] = false;
        for(int i=head[u], v, w; ~i; i=edge[i].nex)
        {
            v = edge[i].to; w = edge[i].val;
            if(dist[v] > dist[u] + w)
            {
                dist[v] = dist[u] + w;
                if(!inque[v])
                {
                    if(++used[v] >= N) return -1;
                    inque[v] = true;
                    Q.push(v);
                }
            }
        }
    }
    return dist[ed];
}
inline void init()
{
    cnt = 0;
    memset(head, -1, sizeof(head));
}
int main()
{
    int T; scanf("%d", &T);
    for(int Cas=1; Cas<=T; Cas++)
    {
        scanf("%d%d", &N, &D);
        init();
        for(int i=1; i<=N; i++)
        {
            if(i > 1) addEddge(i, i-1, -1);
            scanf("%d", &a[i].val);
            a[i].id = i;
        }
        sort(a + 1, a + N + 1, cmp);
        for(int i=2, x1, x2; i<=N; i++)
        {
            x1 = a[i-1].id; x2 = a[i].id;
            if(x1 > x2) swap(x1, x2);
            addEddge(x1, x2, D);
        }
        if(a[1].id > a[N].id) swap(a[1].id, a[N].id);
        printf("Case %d: %d\n", Cas, spfa(a[1].id, a[N].id));
    }
    return 0;
}