题目链接
先\(Tarjan\)缩点，记录每个环内的最大值和最小值。
然后跑拓扑排序，\(Min[u]\)表示到\(u\)的最小值，\(ans[u]\)表示到\(u\)的答案，\(Min\)和\(ans\)都在拓扑排序中更新和传递。
最终答案就是\(ans[n]\)。
\(100\)多行敲着心累
#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;
}