Noip2017Day1T3 逛公园

题目链接

problem

一个有向无重边自环图，设\(D\)为从\(1\)号点走到\(n\)号点的最短距离。问有多少条从\(1\)到\(n\)的路径长度不超过\(D+K\)。\(K\)为给定的值，且\(K\le 50\)

如果有无数条，输出-1

solution

下面有\(dis[i]\)表示\(i\)号点到\(n\)号点的最短路径长度。

设\(f[i][j]\)表示从\(i\)号点走到\(n\)号点，走了\(j\)的多余路径的方案数。就有如下转移:

\[f[i][j]=\sum\limits_{i,v之间有边}f[v][j-(dis[v]+w-dis[i])]\]

记忆化搜索即可。

注意到如果出现了无数条路径，肯定出现了0环。也就是某一个状态被访问了两次。记忆化搜索的过程中标记一下即可。

code

#include<cstdio>
#include<iostream>
#include<ctime>
#include<queue>
#include<cstring>
#include<string>
using namespace std;
typedef long long ll;
const int N = 200010;
ll read() {
    ll x = 0,f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        x = x * 10 + c - '0';
        c = getchar();
    }
    return x * f;
}
struct node {
    int v,nxt,w;
}e[N << 1],E[N << 1];
int head[N],ejs;
void add(int u,int v,int w) {
    e[++ejs].v = v;e[ejs].nxt = head[u];head[u] = ejs;e[ejs].w = w;
}
int head2[N],ejs2;
void add2(int u,int v,int w) {
    E[++ejs2].v = v;E[ejs2].w = w;E[ejs2].nxt = head2[u];head2[u] = ejs2;
}
queue<int>q;
int n,m,K,mod,dis[N],vis[N];
void spfa(int U) {
    memset(dis,0x3f,sizeof(dis));
    memset(vis,0,sizeof(vis));
    dis[U] = 0;
    q.push(U);
    while(!q.empty()) {
        int u = q.front();q.pop();vis[u] = 0;
        for(int i = head2[u];i;i = E[i].nxt) {
            int v = E[i].v;
            if(dis[v] > dis[u] + E[i].w) {
                dis[v] = dis[u] + E[i].w;
                if(!vis[v]) {
                    vis[v] = 1;q.push(v);
                }
            }
        }
    }
}
int bz[N][60],f[N][60];
int dfs(int u,int x) {
    if(bz[u][x] == 2) return f[u][x];
    if(bz[u][x] == 1) return -1;
    bz[u][x] = 1;
    for(int i = head[u];i;i = e[i].nxt) {
        int v = e[i].v;
        int w = x - (dis[v] + e[i].w - dis[u]);
        if(w < 0 || w > K) continue;
        int k = dfs(v,w);
        if(k == -1) return -1;
        f[u][x] += k;
        f[u][x] %= mod;
    }
    bz[u][x] = 2;
    return f[u][x];
} 
int main() {
    int T = read();
    while(T--) {
        memset(head,0,sizeof(head));
        ejs2 = 0;
        memset(head2,0,sizeof(head2));
        ejs = 0;
        n = read(),m = read(),K = read(),mod = read();
        
        memset(f,0,sizeof(f));
        memset(bz,0,sizeof(bz));
        
        for(int i = 1;i <= m;++i) {
            int u = read(),v = read(),w = read();
            add(u,v,w);
            add2(v,u,w);
        }
        
        spfa(n);
        
        f[n][0] = 1;
        int ans = 0;
        for(int i = 0;i <= K;++i) {
            int k = dfs(1,i);
            if(k == -1) {
                ans = -1;break;
            }
            ans += k;
            ans %= mod;
        }
        printf("%d\n",ans);
    }
    
        
    return 0;
}