换句话说，就是寻找一条从1到n的路径，使路径上两点x, y（先经过x再经过y）使val[x] - val[y]最大。
还可以用dp的思想来做这道题：令dis1[x]表示1到x的所有路径中val最小的点，dis2[x]表示从t到x的所有路径中val最大的点，这样答案就是max(dis2[x] - dis1[x])。
用dijkstra就可以实现这个dp。
1 #include<cstdio> 2 #include<iostream> 3 #include<cmath> 4 #include<algorithm> 5 #include<cstring> 6 #include<cstdlib> 7 #include<cctype> 8 #include<vector> 9 #include<stack> 10 #include<queue> 11 using namespace std; 12 #define enter puts("") 13 #define space putchar(' ') 14 #define Mem(a) memset(a, 0, sizeof(a)) 15 typedef long long ll; 16 typedef double db; 17 const int INF = 0x3f3f3f3f; 18 const db eps = 1e-8; 19 const int maxn = 1e5 + 5; 20 inline ll read() 21 { 22 ll ans = 0; 23 char ch = getchar(), last = ' '; 24 while(!isdigit(ch)) {last = ch; ch = getchar();} 25 while(isdigit(ch)) {ans = ans * 10 + ch - '0'; ch = getchar();} 26 if(last == '-') ans = -ans; 27 return ans; 28 } 29 inline void write(ll x) 30 { 31 if(x < 0) x = -x, putchar('-'); 32 if(x >= 10) write(x / 10); 33 putchar(x % 10 + '0'); 34 } 35 36 int n, m, a[maxn]; 37 vector<int> v[maxn], v2[maxn]; 38 39 #define pr pair<int, int> 40 #define mp make_pair 41 int dis1[maxn]; 42 bool in[maxn]; 43 void dij1(int s) 44 { 45 for(int i = 1; i <= n; ++i) dis1[i] = INF; 46 dis1[s] = a[s]; 47 priority_queue<pr, vector<pr>, greater<pr> > q; q.push(mp(dis1[s], s)); 48 while(!q.empty()) 49 { 50 int now = q.top().second; q.pop(); 51 if(in[now]) continue; 52 in[now] = 1; 53 for(int i = 0; i < (int)v[now].size(); ++i) 54 { 55 if(dis1[v[now][i]] > min(dis1[now], a[v[now][i]])) 56 { 57 dis1[v[now][i]] = min(dis1[now], a[v[now][i]]); 58 q.push(mp(dis1[v[now][i]], v[now][i])); 59 } 60 } 61 } 62 } 63 64 int dis2[maxn]; 65 void dij2(int s) 66 { 67 Mem(in); 68 dis2[s] = a[s]; 69 priority_queue<pr> q; q.push(mp(dis1[s], s)); 70 while(!q.empty()) 71 { 72 int now = q.top().second; q.pop(); 73 if(in[now]) continue; 74 in[now] = 1; 75 for(int i = 0; i < (int)v2[now].size(); ++i) 76 { 77 if(dis2[v2[now][i]] < max(dis2[now], a[v2[now][i]])) 78 { 79 dis2[v2[now][i]] = max(dis2[now], a[v2[now][i]]); 80 q.push(mp(dis2[v2[now][i]], v2[now][i])); 81 } 82 } 83 } 84 } 85 86 int ans = 0; 87 88 int main() 89 { 90 n = read(); m = read(); 91 for(int i = 1; i <= n; ++i) a[i] = read(); 92 for(int i = 1; i <= m; ++i) 93 { 94 int x = read(), y = read(), z = read(); 95 v[x].push_back(y); v2[y].push_back(x); 96 if(z == 2) v[y].push_back(x), v2[x].push_back(y); 97 } 98 dij1(1); 99 dij2(n); 100 for(int i = 1; i <= n; ++i) ans = max(ans, dis2[i] - dis1[i]); 101 write(ans); enter; 102 return 0; 103 }