HDU House Man

传送门

思路

很显然，第\(k\)高的必须向第\(k+1\)高的连一条\(D\)的边权
因为所有房屋的顺序不能改变那就是\(x_{i+1} - x_{i} >= 1\)
即\(x_{i } - x_{i+1} <= -1\) 也就是\(i+1\)向\(i\)连一条\(-1\)的边

code

#include <bits/stdc++.h>
#define ll long long
#define N 1010
#define M 1010

using namespace std;
struct node {
	int next, to, dis;
	void clear() {
		next = to = dis = 0;
	}
}edge[N << 2];
struct PPP {
	int x, id, bian;
}point[M];
int add_edge, head[N << 2];
int n, m, st, ed, vis[M], pts[M], dis[M];

int read() {
	int s = 0, f = 0; char ch = getchar();
	while (!isdigit(ch)) f |= (ch == '-'), ch = getchar();
	while (isdigit(ch)) s = s * 10 + (ch ^ 48), ch = getchar();
	return f ? -s : s;
}

void init() {
	for (int i = 1; i <= add_edge; i++)	edge[i].clear();
	add_edge = 0; memset(head, 0, sizeof head);
}
bool cmp(PPP a, PPP b) {
	return a.x < b.x;
}
void add(int from, int to, int dis) {
	edge[++add_edge].next = head[from];
	edge[add_edge].to = to;
	edge[add_edge].dis = dis;
	head[from] = add_edge;
}
int spfa(int st) {
	memset(vis, 0, sizeof vis);
	memset(pts, 0, sizeof pts);
	memset(dis, 0x3f, sizeof dis);
	queue<int> q; q.push(st); 
	dis[st] = 0, vis[st] = 1;
	while (!q.empty()) {
		int fr = q.front(); 
		q.pop(), vis[fr] = 0;
		for (int i = head[fr]; i; i = edge[i].next) {
			int to = edge[i].to;
			if (dis[to] > dis[fr] + edge[i].dis) {
				dis[to] = dis[fr] + edge[i].dis;
				if (!vis[to]) {
					pts[to]++;
					if (pts[to] == n) return -1;
					vis[to] = 1, q.push(to);
				}
			}
		}
	} 
	return dis[ed];
}

int main() {
	int T = read(), ss = T; 
	while (T--) {
		init();
		n = read(), m = read();
		for (int i = 1; i <= n; i++) {
			point[i].id = i, point[i].x = read();
			if (i != n) add(i + 1, i, -1);
		}
		sort(point + 1, point + n + 1, cmp);
		for (int i = 1; i < n; i++) {
			int x = point[i].id, y = point[i + 1].id;
			if (x > y) swap(x, y);
			add(x, y, m);
		}
		st = point[1].id, ed = point[n].id;
		if (st > ed) swap(st, ed);
		int sy = spfa(st);
		printf("Case %d: %d\n", ss - T, sy);
	}
}