差分约束 : 排队布局

排队布局

分析

• \(第一问问是否存在负环,将n个点全放进队列即可\)
• \(如何判断1到N的距离可以无限大?\)
  • \(即判断1到N是否连通.\)

约束关系

• \(x_b\le x_a+L\)
• \(x_b-x_a\ge D\Rightarrow x_a\le x_b-D\)
• \(x_i\le x_{i+1}\)

代码

#include <bits/stdc++.h>
using namespace std;
#define IO ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
inline int lowbit(int x) { return x & (-x); }
#define ll long long
#define pb push_back
#define PII pair<int, int>
#define x first
#define y second
#define inf 0x3f3f3f3f
const int N = 1010, M = 10000 + 10000 + 1000 + 10;
int n, m1, m2;
int h[N], e[M], w[M], ne[M], idx;
int dist[N];
int q[N], cnt[N];
bool st[N];

void add(int a, int b, int c) {
    e[idx] = b, ne[idx] = h[a], w[idx] = c, h[a] = idx++;
}

bool spfa(int size) {
    int hh = 0, tt = 0;
    memset(dist, 0x3f, sizeof dist);
    memset(st, 0, sizeof st);
    memset(cnt, 0, sizeof cnt);
    for (int i = 1; i <= size; ++i) {
        q[tt++] = i;
        dist[i] = 0;
        st[i] = true;
    }
    while (hh != tt) {
        int t = q[hh++];
        if (hh == N) hh = 0;
        st[t] = false;
        for (int i = h[t]; ~i; i = ne[i]) {
            int j = e[i];
            if (dist[j] > dist[t] + w[i]) {
                dist[j] = dist[t] + w[i];
                cnt[j] += 1;
                if (cnt[j] >= n) return true;
                if (!st[j]) {
                    q[tt++] = j;
                    if (tt == N) tt = 0;
                    st[j] = true;
                }
            }
        }
    }
    return false;
}

int main() {
    IO;
    cin >> n >> m1 >> m2;
    memset(h, -1, sizeof h);
    for (int i = 1; i < n; ++i) add(i + 1, i, 0);
    while (m1--) {
        int a, b, c;
        cin >> a >> b >> c;
        if (a > b) swap(a, b);
        add(a, b, c);
    }
    while (m2--) {
        int a, b, c;
        cin >> a >> b >> c;
        if (a > b) swap(a, b);
        add(b, a, -c);
    }
    if (spfa(n)) puts("-1");
    else {
        spfa(1);
        if (dist[n] == inf) puts("-2");
        else cout << dist[n] << '\n';
    }
    return 0;
}