题解集合

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知识点 dp + 最短路

题解

cpp代码

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

using namespace std;

const int N = 100010, M = 2000010;

int n, m;
int w[N];
int hs[N], ht[N], e[M], ne[M], idx;
int dmin[N], dmax[N];
int q[N];
bool st[N];

void add(int h[], int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++ ;
}

void spfa(int h[], int dist[], int type)
{
    int hh = 0, tt = 1;
    if (type == 0)
    {
        memset(dist, 0x3f, sizeof dmin);
        dist[1] = w[1];
        q[0] = 1;
    }
    else
    {
        memset(dist, -0x3f, sizeof dmax);
        dist[n] = w[n];
        q[0] = n;
    }

    while (hh != tt)
    {
        int t = q[hh ++ ];
        if (hh == N) hh = 0;

        st[t] = false;
        for (int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if (type == 0 && dist[j] > min(dist[t], w[j]) || type == 1 && dist[j] < max(dist[t], w[j]))
            {
                if (type == 0) dist[j] = min(dist[t], w[j]);
                else dist[j] = max(dist[t], w[j]);

                if (!st[j])
                {
                    q[tt ++ ] = j;
                    if (tt == N) tt = 0;
                    st[j] = true;
                }
            }
        }
    }
}

int main()
{
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i ++ ) scanf("%d", &w[i]);

    memset(hs, -1, sizeof hs);
    memset(ht, -1, sizeof ht);

    while (m -- )
    {
        int a, b, c;
        scanf("%d%d%d", &a, &b, &c);
        add(hs, a, b), add(ht, b, a);
        if (c == 2) add(hs, b, a), add(ht, a, b);
    }

    spfa(hs, dmin, 0);
    spfa(ht, dmax, 1);

    int res = 0;
    for (int i = 1; i <= n; i ++ ) res = max(res, dmax[i] - dmin[i]);

    printf("%d\n", res);

    return 0;
}