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 "fish.h"
#include <bits/stdc++.h>
using namespace std;
long long max_weights(int N, int M, std::vector<int> X, std::vector<int> Y, std::vector<int> W) {
// first subtask
bool s1 = true, s2=true, s3 = true;
for (int i=0;i<M;i++) {
if (X[i] % 2 != 0) s1=false;
if (X[i] > 1) s2=false;
if (Y[i] != 0) s3 = false;
}
if (s1) {
long long sm = 0;
for (int i=0;i<M;i++) sm += W[i];
return sm;
}
if (s2) {
long long sm1=0,sm2=0;
for (int i=0;i<M;i++) {
if (X[i] == 0) sm1 += W[i];
else sm2 += W[i];
}
if (N == 2) return max(sm1, sm2);
vector<long long> p(N, 0), s(N, 0);
for (int i=0;i<M;i++) {
if (X[i] == 0) p[Y[i]] += W[i];
else s[Y[i]] += W[i];
}
for (int i=1;i<N;i++) p[i]+=p[i-1];
for (int i=N-2;i>=0;i--) s[i] += s[i+1];
long long mx = max(p[N-1], s[0]);
for (int i=1;i<N;i++) {
mx = max(mx, p[i-1]+s[i]);
}
return mx;
}
if (s3) {
vector<int> w(N,0);
for (int i=0;i<M;i++) {
w[X[i]] = W[i];
}
vector<long long> dp1(N, 0), dp2(N, 0), dp3(N, 0);
dp1[0] = dp2[0] = dp3[0] = 0;
for (int i=1;i<N;i++) {
// build smt
dp1[i] = max(dp1[i-1], max(dp2[i-1], dp3[i-1] + w[i-1]));
dp2[i] = max(dp1[i-1] + w[i], max(dp2[i-1], dp3[i-1]));
dp3[i] = max(dp3[i-1], dp2[i-1]);
}
return max(dp1.back(), max(dp2.back(), dp3.back()));
}
vector<vector<long long>> dp1(N, vector<long long>(N+1, 0)), dp2(N, vector<long long>(N+1, 0));
// dp1[i][j] : maximum number of fish such that row i has pier with height j
// dp2[i][j] : maximum number of fish such that row i has pier with height j and row i-1 has pier with height no greater than j
vector<vector<long long>> s(N, vector<long long>(N+1, 0));
for (int i=0;i<M;i++) {
s[X[i]][Y[i]+1] = W[i];
}
for (int i=N-1;i>=0;i--) {
for (int j=N-1;j>=0;j--) {
s[i][j] += s[i][j+1];
}
}
long long mx = 0;
for (int i=1;i<N;i++) {
for (int j=0;j<=N;j++) {
long long sub = (j == N ? 0 : s[i-1][j+1]);
for (int k=0;k<=j;k++) {
dp1[i][j] = max(dp1[i][j], max(dp2[i-1][k] + (k == N ? 0 : s[i-1][k+1]) - sub, dp1[i-1][k]));
}
dp2[i][j] = dp1[i][j];
sub = (j == N ? 0 : s[i][j+1]);
for (int k=j+1;k<=N;k++) {
dp1[i][j] = max(dp1[i][j], max(dp1[i-1][k], dp2[i-1][k]) + sub - (k == N ? 0 : s[i][k+1]));
}
mx = max(mx, max(dp1[i][j], dp2[i][j]));
}
}
return mx;
}
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