題目鏈接
先\(Tarjan\)縮點，記錄每個環內的最大值和最小值。
然後跑拓撲排序，\(Min[u]\)表示到\(u\)的最小值，\(ans[u]\)表示到\(u\)的答案，\(Min\)和\(ans\)都在拓撲排序中更新和傳遞。
最終答案就是\(ans[n]\)。

\(100\)多行敲著心累

#include <cstdio>
#include <cstring>
#define Open(s) freopen(s".in","r",stdin);freopen(s".out","w",stdout);
#define Close fclose(stdin);fclose(stdout);
namespace IO{
    int xjc; char ch;
    inline int read(){
        xjc = 0; ch = getchar();
        while(ch < '0' || ch > '9') ch = getchar();
        while(ch >= '0' && ch <= '9'){ xjc = xjc * 10 + ch - '0'; ch = getchar(); }
        return xjc;
    }
}using namespace IO;
inline int max(int a, int b){
    return a > b ? a : b;
}
inline int min(int a, int b){
    return a > b ? b : a;
}
const int MAXN = 100010;
const int MAXM = 500010;
struct Queue{
    int s[MAXN];
    int head, tail;
    inline void push(int x){
        s[++tail] = x;
    }
    inline int pop(){
        return s[++head];
    }
    inline int size(){
        return tail - head;
    }
}q;
struct Edge{
    int from, next, to;
};
struct Graph{
   int head[MAXN], num;
   Edge e[MAXM << 1];
   inline void Add(int from, int to){
    e[++num].to = to; e[num].from = from; e[num].next = head[from]; head[from] = num;
   }
}G, T, P;
int dfn[MAXN], low[MAXN], id, vis[MAXN], stack[MAXN], w[MAXN], top, cnt, belong[MAXN], v[MAXN];
int minw[MAXN], maxw[MAXN], now, in[MAXN], Min[MAXN], ans[MAXN];
void Tarjan(int u){
    dfn[u] = low[u] = ++id; vis[u] = 1; stack[++top] = u;
    for(int i = G.head[u]; i; i = G.e[i].next)
       if(!dfn[G.e[i].to]){
         Tarjan(G.e[i].to);
         low[u] = min(low[u], low[G.e[i].to]);
       }
       else if(vis[G.e[i].to])
         low[u] = min(low[u], dfn[G.e[i].to]);
    if(dfn[u] == low[u]){
      ++cnt;
      do{
        now = stack[top--];
        vis[now] = 0;
        belong[now] = cnt;
        minw[cnt] = min(minw[cnt], w[now]);
        maxw[cnt] = max(maxw[cnt], w[now]);
        P.Add(cnt, now);
      }while(now != u);
    }
}
int n, m, a, b, c;
int main(){
    Open("trade");
    memset(minw, 127, sizeof minw);
    memset(Min, 127, sizeof Min);
    n = read(); m = read();
    for(int i = 1; i <= n; ++i)
       w[i] = read();
    for(int i = 1; i <= m; ++i){
       a = read(); b = read(); c = read();
       G.Add(a, b);
       if(c == 2) G.Add(b, a);
    }
    for(int i = 1; i <= n; ++i)
       if(!dfn[i])
         Tarjan(i);
    for(int i = 1; i <= cnt; ++i){
       for(int j = P.head[i]; j; j = P.e[j].next){
          int u = P.e[j].to;
          for(int k = G.head[u]; k; k = G.e[k].next)
             if(belong[G.e[k].to] != i && !v[belong[G.e[k].to]]){
               v[belong[G.e[k].to]] = 1;
               T.Add(i, belong[G.e[k].to]);
               ++in[belong[G.e[k].to]];
             }
       }
       for(int j = P.head[i]; j; j = P.e[j].next){
          int u = P.e[j].to;
          for(int k = G.head[u]; k; k = G.e[k].next)
             v[belong[G.e[k].to]] = 0;
       }
    }
    q.push(belong[1]);
    while(q.size()){
      now = q.pop();
      Min[now] = min(Min[now], minw[now]);
      ans[now] = max(ans[now], maxw[now] - Min[now]);
      for(int i = T.head[now]; i; i = T.e[i].next){
         int v = T.e[i].to;
         if(!(--in[v])) q.push(v);
         Min[v] = min(Min[v], Min[now]);
         ans[v] = max(ans[v], ans[now]);
      }
    }
    printf("%d\n", ans[belong[n]]);
    return 0;
}
 

【洛谷 P1073】 最優貿易 （Tarjan縮點+拓撲排序）