同余最短路

主要是解决几个数能否拼凑成其他数的问题

给定三个数 $x,y,z$ ,问有多少 $w=ax+by+cz\le h-1$

定义 $dis_i$ 表示 $x,y,z$ 能够组成 $w\mod x=i$ 的最小的数

可以这样建图 $i\stackrel{y}{\longrightarrow}i+y\mod x,i\stackrel{z}{\longrightarrow}i+z\mod x$

这样就只需要跑一遍最短路就可以求出 $dis$ 了,统计答案时若 $dis_i\le h-1$ ,就加上 $\lfloor\frac{h-1-dis_i}{x}\rfloor +1$

代码

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;
const int N = 2e5 + 5;
typedef long long ll;
typedef pair<ll, int> PII;

ll dis[N], edge[N];
int head[N], ver[N], net[N], idx;
priority_queue<PII, vector<PII>, greater<PII> > q;
bool vis[N];

void add(int a, int b, int c)
{
    net[++idx] = head[a], ver[idx] = b, edge[idx] = c, head[a] = idx;
}

void Dij()
{
    memset(dis, 0x3f, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    q.push(make_pair(0, 0));
    dis[0] = 0;
    while (!q.empty())
    {
        int u = q.top().second;
        q.pop();
        if (vis[u])
            continue;
        vis[u] = true;
        for (int i = head[u]; i; i = net[i])
        {
            int v = ver[i];
            if (dis[v] > dis[u] + edge[i])
            {
                dis[v] = dis[u] + edge[i];
                q.push(make_pair(dis[v], v));
            }
        }
    }
}

int main()
{
    ll h;
    int x, y, z;
    scanf("%lld%d%d%d", &h, &x, &y, &z);
    h--;
    for (int i = 0; i < x; i++)
        add(i, (i + y) % x, y), add(i, (i + z) % x, z);
    Dij();
    ll ans = 0;
    for (int i = 0; i < x; i++)
        if (dis[i] <= h)
            ans += (h - dis[i]) / x + 1;
    printf("%lld", ans);
    return 0;
}

练习P2371 [国家集训队]墨墨的等式

和上面的一样,就直接放代码了

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int N = 6e6 + 5;
typedef pair<int, int> PII;
typedef long long ll;

int a[15];
ll dis[N];
int head[N], net[N], ver[N], edge[N], idx;
priority_queue<PII, vector<PII>, greater<PII> > q;
bool vis[N];

void add(int a, int b, int c)
{
    net[++idx] = head[a], ver[idx] = b, edge[idx] = c, head[a] = idx;
}

void Dij()
{
    memset(dis, 0x3f, sizeof(dis));
    memset(vis, 0, sizeof(vis));
    dis[0] = 0;
    q.push(make_pair(0, 0));
    while (!q.empty())
    {
        int u = q.top().second;
        q.pop();
        if (vis[u])
            continue;
        vis[u] = true;
        for (int i = head[u]; i; i = net[i])
        {
            int v = ver[i];
            if (dis[v] > dis[u] + edge[i])
            {
                dis[v] = dis[u] + edge[i];
                q.push(make_pair(dis[v], v));
            }
        }
    }
}

int main()
{
    int n;
    ll l, r;
    scanf("%d%lld%lld", &n, &l, &r);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    for (int i = 0; i < a[1]; i++)
        for (int j = 2; j <= n; j++)
            add(i, (i + a[j]) % a[1], a[j]);
    Dij();
    ll ans = 0;
    for (int i = 0; i < a[1]; i++)
    {
        if (dis[i] <= r)
            ans += (r - dis[i]) / a[1] + 1;
        if (dis[i] < l)
            ans -= (l - 1 - dis[i]) / a[1] + 1;
    }
    printf("%lld", ans);
    return 0;
}