41 lines
1.0 KiB
C++
41 lines
1.0 KiB
C++
// P4644
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#include <bits/stdc++.h>
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struct node
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{
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int l, r, s;
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} a[10010];
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int n, m, e;
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using namespace std;
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bool cmp(node x, node y)
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{
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return x.r < y.r;
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}
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long long f[86400];
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int main()
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{
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// freopen("cleaning.in", "r", stdin);
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// freopen("cleaning.out", "w", stdout);
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scanf("%d%d%d", &n, &m, &e);
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++m, ++e;
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memset(f, 0x7f, sizeof(f));
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f[m - 1] = 0;
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for (int i = 0; i < n; ++i)
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{
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scanf("%d%d%d", &a[i].l, &a[i].r, &a[i].s);
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++a[i].l, ++a[i].r;
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}
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sort(a, a + n, cmp);
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for (int i = 0; i < n; ++i)
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{
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long long minfj = 0x7f7f7f7f7f7f7f7f;
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for (int j = max(m - 1, a[i].l - 1); j <= min(e, a[i].r - 1); ++j)
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if (f[j] < minfj)
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minfj = f[j];
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f[min(a[i].r, e)] = min(f[min(a[i].r, e)], minfj + a[i].s);
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}
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printf("%lld\n", (f[e] == 0x7f7f7f7f7f7f7f7f) ? -1 : f[e]);
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return 0;
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}
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// 设f[x]为清理[L,x]需要花费的最小代价
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// f[min(a[i].r,e)]=min(f[j])+a[i].s (max(L-1,a[i].l-1)<=j<=min(R,c[i].r-1))
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