79 lines
1.8 KiB
C++
79 lines
1.8 KiB
C++
#include <bits/stdc++.h>
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using namespace std;
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int n, m, k;
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long long x, y;
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int a[200005];
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int b, preb;
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unordered_map<int, int> wz;
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bool k_is_better;
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vector<int> mor;
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long long del(int pre, int todo)
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{
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if (todo <= pre)
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return -1;
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if (todo == pre + 1)
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return 0;
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mor.clear();
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for (int j = pre + 1; j <= todo - 1; ++j)
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if (a[j] > max(a[todo], a[pre]))
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mor.push_back(j);
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if (!mor.empty() && (todo - pre - 1) < k) // 如果有大数且长度不足以运行一次 k 则不存在
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return -1;
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if (k_is_better)
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{
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return ((todo - pre - 1) % k) * y + ((todo - pre - 1) / k) * x; // 先用 y 移除多余的数，最后用多个 k 直接删完
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}
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else
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{
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if (!mor.empty())
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return (todo - pre - 1 - k) * y + x; // 用一次k清除最大数，其余用y
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else
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return (todo - pre - 1) * y; // 全部用y
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}
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return 0;
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}
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int main()
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{
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scanf("%d%d%lld%d%lld", &n, &m, &x, &k, &y);
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k_is_better = x < (k * y);
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for (int i = 1; i <= n; ++i)
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{
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scanf("%d", &a[i]);
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wz[a[i]] = i;
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}
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long long ans = 0;
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for (int i = 1; i <= m; ++i)
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{
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scanf("%d", &b);
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int todo = wz[b], pre = 0;
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if (i != 1)
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pre = wz[preb];
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long long d = del(pre, todo);
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if (d == -1ll)
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{
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printf("-1\n");
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return 0;
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}
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ans += d;
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preb = b;
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}
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long long d = del(wz[b], n + 1);
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if (d == -1ll)
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{
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printf("-1\n");
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return 0;
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}
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printf("%lld\n", ans + d);
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return 0;
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}
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/*
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若 k 删除更优，先删到能被 k 整除的长度，则先 k 删除，最后运行多次 k 删除
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len % k
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若 1 删除更优，先用 k 删除删去大于两边的，最后用 1 删除
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7 2
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3 3 1
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5 7 0 3 1 6 4
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5 4
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输出 5
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*/ |