传送门

坎坷经历（看题解的可略过）

• 其实这道题还是有点意思的，，，
• 其实看到这道题我脑子里想的一直是Tarjan缩点然后DAGdp，也不知道能不能做
• 看了眼题解好吧，，两遍SPFA，都是套路，，，
• 正经八本地打完SPFA发现不对劲，如果按照常规的SPFA开始加入的节点只有1号点，再加入其它节点时会发生问题，，
• 正经八本地把所有点都加进去，发现有些点其实是访问不到的，不能用它们更新答案
• 最后改对了。。

正经的题解

• 记minn和maxx数组，保存从1开始到x的最小值以及从n开始到x的最大值
• 把minn数组初始化为INF，maxx初始化为0
• 从1开始SPFA，把所有能更新的和值为INF的进行更新，注意，先考虑更新INF，在考虑松弛操作
• 从n开始反向SPFA
• 注意一下建边的时候要建正向边和反向边
• 枚举一遍1到n，根据$(maxx[i] - minn[i])$更新ans，输出ans即为结果

#include <cstdio>
#include <cstring>
#include <algorithm>

const int maxm = 2000000 + 50000;
const int maxn = 150000;
int minn[maxn], maxx[maxn];
int last[maxn], pre[maxm], other[maxm], len[maxm];
int tot = 0;
int n, m;
int x, y, z;
int que[maxn], vis[maxn];
int a[maxn];

void add(int x, int y, int z) {
    tot++;
    pre[tot] = last[x];
    last[x] = tot;
    other[tot] = y;
    len[tot] = z;
}

void spfa1(void) {
    que[1] = 1;
    minn[1] = a[1];
    int queh = 0, quet = 1;
    while (queh != quet) {
        queh = (queh + 1) % maxn;
        int cur = que[queh];
        vis[cur] = 0;
        for (int p = last[cur]; p; p = pre[p]) {
            if (len[p] == 2) continue;
            int q = other[p];
            if (minn[q] == 2139062143) {
                minn[q] = a[q];
                vis[q] = 1;
                quet = (quet + 1) % maxn;
                que[quet] = q;
            }
            if (minn[q] > minn[cur]) {
                minn[q] = minn[cur];
                if (!vis[q]) {
                    vis[q] = 1;
                    quet = (quet + 1) % maxn;
                    que[quet] = q;
                }
            }
        }
    } 
}

void spfa2(void) {
    que[1] = n;
    maxx[n] = a[n];
    int queh = 0, quet = 1;
    while (queh != quet) {
        queh = (queh + 1) % maxn;
        int cur = que[queh];
        vis[cur] = 0;
        for (int p = last[cur]; p; p = pre[p]) {
            if (len[p] == 1) continue;
            int q = other[p];
            if (maxx[q] == 0) {
                maxx[q] = a[q];
                quet = (quet + 1) % maxn;
                vis[q] = 1;
                que[quet] = q;
            }
            if (maxx[q] < maxx[cur]) {
                maxx[q] = maxx[cur];
                if (!vis[q]) {
                    vis[q] = 1;
                    quet = (quet + 1) % maxn;
                    que[quet] = q;
                }
            }

        }
    } 
}

int main () {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int i = 1; i <= m; i++) {
        scanf("%d %d %d", &x, &y, &z);
        if (z == 2) {
            add(x, y, 1);
            add(y, x, 1);
            add(x, y, 2);
            add(y, x, 2);
        } else {
            add(x, y, 1);
            add(y, x, 2);
        }
    }
    memset(minn, 127, sizeof(minn));
    memset(maxx, 0, sizeof(maxx));
    spfa1();
    spfa2();
    int ans = 0;
    for (int i = 1; i <= n; i++) {
        ans = std :: max(ans, maxx[i] - minn[i]);
    }
    printf("%d", ans);
    return 0;
}