new file: P4644.cpp

P4644.cpp

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