【LOJ2316】「NOIP2017」逛公园

【题目链接】

• 【点击打开链接】

【题目概括】

对给定\(K\)，起点\(1\)到终点\(n\)中对长度为\([L,L+K]\)的路径计数。

\(L\)为\(1\)到\(n\)的最短路长度。

【思路要点】

• 考虑\(DP\)，定义\(dp[i][j]\)表示从\(i\)点到终点\(n\)的距离比原来的最短距离长了\(j\)的方案数。
• 那么考虑转移，\(dp[u][j]=\sum dp[v][j-g]\)，其中\(g=dis[v]+w(i,j)-dis[u]\)。

【代码】

#include <bits/stdc++.h>

#define inf 0x3f3f3f3f

using namespace std;

const int N = 2e5 + 5;

struct _edge {
    int to, nt, w;
} E1[N], E2[N];

int H1[N], H2[N], dis[N];
int edgeCnt1, edgeCnt2;
bool vis[N];
bool ban[N][55];
int dp[N][55];
int n, m, k, MOD;

void addEdge1(int u, int v, int w) {
    E1[++edgeCnt1] = (_edge){v, H1[u], w};
    H1[u] = edgeCnt1;
}

void addEdge2(int u, int v, int w) {
    E2[++edgeCnt2] = (_edge){v, H2[u], w};
    H2[u] = edgeCnt2;
}

void init() {
    edgeCnt1 = edgeCnt2 = 0;
    for (int i = 1; i <= n; i++)
        H1[i] = H2[i] = 0;
    for (int i = 1; i <= n; i++)
        for (int j = 0; j <= k; j++)
            dp[i][j] = -1, ban[i][j] = 0;
}

void spfa(int s) {
    for (int i = 1; i <= n; i++)
        dis[i] = inf, vis[i] = 0;
    queue<int> q;
    q.push(s);
    vis[s] = 1;
    dis[s] = 0;
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        vis[u] = 0;
        for (int e = H2[u]; e; e = E2[e].nt) {
            int v = E2[e].to;
            if (dis[v] > dis[u] + E2[e].w) {
                dis[v] = dis[u] + E2[e].w;
                if (!vis[v]) {
                    q.push(v);
                    vis[v] = 1;
                }
            }
        }
    }
}

int dfs(int u, int k) {
    if (k < 0)
        return 0;
    else if (ban[u][k])
        return -1;
    else if (dp[u][k] != -1)
        return dp[u][k];
    ban[u][k] = 1;
    dp[u][k] = 0;
    if (u == n)
        dp[u][k]++;
    for (int e = H1[u]; e; e = E1[e].nt) {
        int v = E1[e].to;
        if (dis[v] == inf)
            continue;
        int g = dis[v] + E1[e].w - dis[u];
        int res = dfs(v, k - g);
        if (res == -1)
            return -1;
        dp[u][k] = (dp[u][k] + res) % MOD;
    }
    ban[u][k] = 0;
    return dp[u][k];
}

int main() {
    int T; scanf("%d", &T);
    while (T--) {
        scanf("%d %d %d %d", &n, &m, &k, &MOD);
        init();
        for (int i = 1; i <= m; i++) {
            int u, v, w; scanf("%d %d %d", &u, &v, &w);
            addEdge1(u, v, w), addEdge2(v, u, w);
        }
        spfa(n);
        int ans = dfs(1, k);
        printf("%d\n", ans);
    }
    return 0;
}