[NOIP2017] 逛公园 【最短路】【强连通分量】

题目分析:

首先考虑无数条的情况。出现这种情况一定是一条合法路径经过了$ 0 $环中的点。那么预先判出$ 0 $环中的点和其与$ 1 $和$ n $的距离。加起来若离最短路径不超过$ k $则输出$ -1 $，否则这些点必定不被经过，接着dp后效性消失。由于每条边转移了$ k $次它的起点到终点的状态，所以总时间复杂度为$ O(n+mk) $，可以通过所有数据。

代码:

 #include<iostream>
 #include<cstdio>
 #include<cstdlib>
 #include<cstring>
 #include<queue>
 #include<vector>
 #include<stack>
 using namespace std; typedef pair<int,int> pr; const int maxn = ; int n,m,k,p;
 vector<pair<int,int> > g[maxn],rg[maxn];
 vector <int> ng[maxn]; int dfn[maxn],low[maxn],instack[maxn],scc[maxn],cl;
 int f[maxn][],arr[maxn][];
 stack <int> sta; void init(){
     memset(dfn,,sizeof(dfn));
     memset(arr,,sizeof(arr));
     memset(low,,sizeof(low));
     memset(instack,,sizeof(instack));
     memset(scc,,sizeof(scc));
     memset(f,,sizeof(f));
     cl = ;
 } void tarjan(int now){
     low[now] = dfn[now] = ++cl;
     sta.push(now);
     for(int i=;i<ng[now].size();i++){
     int k = ng[now][i]; if(instack[k]) continue;
     if(dfn[k]) { low[now] = min(low[now],dfn[k]); }
     else{ tarjan(k); low[now] = min(low[now],low[k]); }
     }
     if(low[now] == dfn[now]){
     stack<int> tpp;int num = ;
     while(true){
         int tp = sta.top();sta.pop();tpp.push(tp);num++;
         instack[tp] = ; scc[tp]=;
         if(now == tp) break;
     }
     if(num == ) scc[tpp.top()] = ;
     }
 } void read(){
     scanf("%d%d%d%d",&n,&m,&k,&p);
     for(int i=;i<=n;i++) g[i].clear();
     for(int j=;j<=n;j++) ng[j].clear();
     for(int j=;j<=n;j++) rg[j].clear();
     for(int i=;i<=m;i++){
     int u,v,w; scanf("%d%d%d",&u,&v,&w);
     g[u].push_back(make_pair(v,w));
     rg[v].push_back(make_pair(u,w));
     if(w == ) ng[u].push_back(v);
     }
 } int res;
 int d[maxn][];
 priority_queue<pr,vector<pr>,greater<pr> > pq;
 void solve_dist(){
     memset(d,0x3f,sizeof(d));
     pq.push(make_pair(,));
     while(!pq.empty()){
     pair<int,int> tp = pq.top();pq.pop();
     if(d[tp.second][] > 1e9) d[tp.second][] = tp.first;
     else continue;
     for(int j=;j<g[tp.second].size();j++){
         int nxt = g[tp.second][j].first,data = g[tp.second][j].second;
         data += d[tp.second][];
         if(data > d[nxt][]) continue;
         pq.push(make_pair(data,nxt));
     }
     }
     res = d[n][];
     pq.push(make_pair(,n));
     while(!pq.empty()){
     pair<int,int> tp = pq.top();pq.pop();
     if(d[tp.second][] > 1e9) d[tp.second][] = tp.first;
     else continue;
     for(int j=;j<rg[tp.second].size();j++){
         int nxt = rg[tp.second][j].first,data = rg[tp.second][j].second;
         data += d[tp.second][];
         if(data > d[nxt][]) continue;
         pq.push(make_pair(data,nxt));
     }
     }
 } void dfs(int now,int lw){
     if(arr[now][lw]) return;
     int nowd = d[now][]+lw;arr[now][lw] = ;
     for(int j=;j<rg[now].size();j++){
     int nxt = rg[now][j].first,data = rg[now][j].second;
     if(scc[nxt])continue;
     int nd = nowd-data; nd -= d[nxt][];
     if(nd < ) continue;
     if(nd > k) continue;
     dfs(nxt,nd);
     f[now][lw] += f[nxt][nd]; f[now][lw] %= p;
     }
 } void work(){
     for(int i=;i<=n;i++) if(!dfn[i]) tarjan(i);
     if(scc[]||scc[n]) {puts("-1");return;}     solve_dist();     for(int i=;i<=n;i++)
     if(scc[i]) if(d[i][]+d[i][]<=res+k){puts("-1");return;}     f[][] = ;arr[][] = ;int ans = ;
     for(int i=;i<=k;i++) {
     dfs(n,i);ans += f[n][i]; ans %= p;
     }
     printf("%d\n",ans);
 } int main(){
     int t; scanf("%d",&t);
     while(t--){ init(); read(); work(); }
     return ;
 }