| # | Time | Username | Problem | Language | Result | Execution time | Memory |
|---|---|---|---|---|---|---|---|
| 896479 | nightfal | Wiring (IOI17_wiring) | C++14 | 0 ms | 0 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
// #include "wiring.h"
#include <cassert>
#include <cstdio>
#include <vector>
#include <climits>
#include <stdlib.h>
#include <algorithm>
using namespace std;
bool isSubtask1(std::vector<int> r, std::vector<int> b) {return r.size()<=200 && b.size()<=200;}
bool isSubtask2(std::vector<int> r, std::vector<int> b) {return r[r.size()-1]<b[0];}
bool isSubtask3(std::vector<int> r, std::vector<int> b) {
int n=r.size(), m=b.size();
vector<int> a(n+m);
for(int i=0, j=0; i<n || j<m;)
i==n? a[i+j++]=1: j==m? a[j+i++]=0: r[i]<b[j]? a[j+i++]=0:a[i+j++] = 1;
int cnt = 1;
for(int k=1; k<n+m; k++) {
cnt = a[k]==a[k-1]? cnt+1:0;
if (cnt > 6) return false;
}
return true;
}
long long subtask1(std::vector<int>, std::vector<int>);
long long subtask2(std::vector<int> &, std::vector<int> &, int, int, int, int);
long long subtask3(std::vector<int>, std::vector<int>);
long long min_total_length(std::vector<int>, std::vector<int> );
int main() {
int n, m;
assert(2 == scanf("%d %d", &n, &m));
vector<int> r(n), b(m);
for(int i = 0; i < n; i++)
assert(1 == scanf("%d", &r[i]));
for(int i = 0; i < m; i++)
assert(1 == scanf("%d", &b[i]));
long long res = min_total_length(r, b);
printf("%lld\n", res);
return 0;
}
long long min_total_length(std::vector<int> r, std::vector<int> b) {
if (isSubtask3(r,b)) return subtask3(r,b);
else if (isSubtask2(r,b)) return subtask2(r,b,0,r.size()-1,0,b.size()-1);
else if (isSubtask1(r,b)) return subtask1(r,b);
// else if (isSubtask4(r,b)) return subtask4(r,b);
// else return fullTask(r,b);
return 0;
}
long long subtask3(std::vector<int> r, std::vector<int> b) {
int n = r.size(), m = b.size();
vector<pair<int,int>> a(n+m);
for(int i=0, j=0; i<n || j<m;) // make_pair 내부에서 ++ 하면 안됨
i==n? a[i+j++]=make_pair(j,1): j==m? a[j+i++]=make_pair(i,0): r[i]<b[j]? a[j+i++]=make_pair(i,0): a[i+j++]=make_pair(j,1);
// for(int i=0; i<a.size(); i++)
// printf("a[%d] = (%d,%d), ", i,a[i].first,a[i].second);
// printf("\n");
vector<long long> dp(n+m,LLONG_MAX);
int prev=0, curr=0;
for(curr=1; curr<n+m; curr++)
if (a[curr].second != a[curr-1].second) break;
// printf("prev: %d, curr: %d\n", prev,curr);
int i;
for(i=curr; i<n+m; i++) {
if (a[prev].second==0)
dp[i] = subtask2(r,b,a[prev].first,a[curr-1].first,a[curr].first,a[i].first);
else
dp[i] = subtask2(b,r,a[prev].first,a[curr-1].first,a[curr].first,a[i].first);
// printf("dp[%d] = %lld\n", i,dp[i]);
if(i<n+m-1 && a[i].second != a[i+1].second) {prev=curr; curr=i+1; break;}
}
if (i==n+m) return dp[n+m-1];
// printf("prev: %d, curr: %d\n", prev,curr);
// for(int i=0; i<curr; i++)
// printf("dp[%d] = %lld\n", i,dp[i]);
for(int i=curr; i<n+m; i++) {
// printf("prev: %d, curr: %d\n", prev,curr);
for(int j =prev; j<curr; j++) {
// printf("i:%d, j:%d\n",i,j);
long long temp;
if (a[prev].second==0)
temp = subtask2(r,b,a[j].first,a[curr-1].first,a[curr].first,a[i].first);
else
temp = subtask2(b,r,a[j].first,a[curr-1].first,a[curr].first,a[i].first);
// printf("temp=%d, ", temp);
if (j==prev) temp += min(dp[j-1],dp[j]);
else temp += dp[j-1];
// printf("temp=%d, ", temp);
if (dp[i]>temp) dp[i] = temp;
// printf("dp[%d] = %lld\n", i,dp[i]);
}
if (i<n+m-1 && a[i].second != a[i+1].second) {prev = curr; curr = i+1;}
}
return dp[n+m-1];
}
long long subtask2(std::vector<int> &arr1, std::vector<int> &arr2, int s1, int e1, int s2, int e2) {
long long sum1=0, sum2=0, n = e1-s1+1, m=e2-s2+1;
for(int i=s1; i<=e1; i++) sum1 += arr1[e1] - arr1[i];
for(int i=s2; i<=e2; i++) sum2 += arr2[i] - arr2[s2];
return sum1+sum2+max(n,m)*(arr2[s2]-arr1[e1]);
}
long long subtask1(std::vector<int> r, std::vector<int> b) {
int n = r.size(), m = b.size();
long long dp[n][m];
dp[0][0] = abs(r[0]-b[0]);
for(int i=1; i<n; i++) dp[i][0] = dp[i-1][0] + abs(r[i]-b[0]);
for(int j=1; j<m; j++) dp[0][j] = dp[0][j-1] + abs(r[0]-b[j]);
for(int i=1; i<n; i++)
for(int j=1; j<m; j++)
dp[i][j] = min(min(dp[i-1][j],dp[i][j-1]),dp[i-1][j-1]) + abs(r[i]-b[j]);
return dp[n-1][m-1];
}