# [CODEFORCES724E] Goods Transportation

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[CODEFORCES724E] Goods Transportation

Question Description

There is N cities located along the one-way road. Cities is numbered from 1 to n in the direction of the road.

The  I -th City had produced  p i  units of goods. No more Than  s i  units of goods can Sold in The  I -th city.

For the pair of cities I and J such that 1?≤? I? < j. ≤? n You can no further than once transport no more than C units of goods by the city C14>i to the city J. Note that goods can is transported from a city with a lesser index to the city with a larger index. You can transport goods between cities on any order.

Determine the maximum number of produced goods that can is sold in total in all the cities after a sequence of transportat Ions.

Input

The first line of the input contains the integers n and C (1?≤? N. ≤?10?000, 0?≤? C. ≤?109)-the number of cities and the maximum amount of goods for a single transportation.

The second line contains n integers pi (0?≤? Pi? ≤?109)-the number of units of goods that were produced in each city.

The third line of input contains n integers si (0?≤? si? ≤?109)-the number of units of goods that can is sold in each city.

Output

Print The maximum total number of produced goods so can be sold in all cities after a sequence of transportations.

Input example

`4 3  - Ten 7 4 4 7 Ten  -`

Output example

`34`

Data size and conventions

See " Input "

Exercises

This is obviously a maximum flow model, from the source point to each city with a capacity of one-way pi, from each city to the meeting point a capacity Si one-way side, and then each city to a number than it is larger than all cities with a capacity of a one-way edge of C.

However, there are up to 10,002 points, 50.015 million sides, time aside, each side needs to have a reverse arc, and each arc needs to save at least 2 int (the end of the edge and the remaining flow of the edge), for the 256MB limit is too large.

However, this problem has a better approach, due to the specificity of the map, we can be converted to the minimum cut DP, set F (i, j) to indicate the first I point J points belonging to the S set of the minimum cut is how much.

First step: First of all, from the source to node I and then to the meeting point of the flow is exhausted, so the residual network becomes this way:

If pi > Si, then a pi-si edge is connected from the source point to I;

If pi < Si, a si-pi edge is connected from I to the meeting point;

If 0 < i < J < n + 1, the edge of a C is connected from I to J.

Step two: Consider how the above DP is transferred, and whether the enumeration current point belongs to the S-set or T-set. I am very lazy, do not want to write again, transfer equation may leave the reader to think about it ...

`#include <iostream> #include <cstdio> #include <algorithm> #include <cmath> #include <stack > #include <vector> #include <queue> #include <cstring> #include <string> #include <map > #include <set>using namespace std;const int buffersize = 1 << 16;char buffer[buffersize], *head, *TAIL;INL        ine Char Getchar () {if (Head = = Tail) {int L = fread (buffer, 1, buffersize, stdin);    Tail = (Head = buffer) + L; } return *head++;}    int read () {int x = 0, f = 1; char c = Getchar ();    while (!isdigit (c)) {if (c = = '-') f =-1; c = Getchar ();}    while (IsDigit (c)) {x = x * + C-' 0 '; c = Getchar ();} return x * f;}  #define MAXN 10010#define oo (1ll <<) #define LL long Longll f[2][maxn];int N, C, P[MAXN], S[maxn];int Main () {n = Read ();  C = read (); for (int i = 1; I <= n; i++) P[i] = Read (), for (int i = 1; I <= n; i++) s[i] = Read (), for (int i = 0; i < 2; i++) for (int j = 0; J <= N; j + +) F[i][j] = Oo;f[0] [0] = 0; LL base = 0; BOOL cur = 1;for (int i = 1; I <= n; i++, cur ^= 1) {base + = min (S[i], p[i]); for (int j = 0; J <= N; j + +) F[cur][j] = oo;for (int j = 0; J <= I; j + +) {if (S[i] > P[i]) {if (J && f[cur^1][j-1] < oo) f[cur][j] = min (F[cur][j], F [Cur^1] [j-1] + s[i]-p[i]), if (F[cur^1][j] < oo) f[cur][j] = min (F[cur][j], F[cur^1][j] + (LL) J * C);}  else {if (J && f[cur^1][j-1] < oo) f[cur][j] = min (F[cur][j], f[cur^1][j-1]), if (F[cur^1][j] < oo) F[cur][j] = Min (F[cur][j], f[cur^1][j] + p[i]-s[i] + (LL) J * C);}} LL tmp = oo;for (int i = 0; I <= n; i++) tmp = MIN (tmp, f[cur^1][i]);p rintf ("%i64d\n", base + tmp); return 0;}`

[CODEFORCES724E] Goods Transportation

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