#include <bits/stdc++.h> using namespace std; #define REP(i,a,b) for (int i = (a); i <= (b); ++i) #define REPD(i,a,b) for (int i = (a); i >= (b); --i) #define FORI(i,n) REP(i,1,n) #define FOR(i,n) REP(i,0,int(n)-1) #define mp make_pair #define pb push_back #define pii pair<int,int> #define vi vector<int> #define ll long long #define SZ(x) int((x).size()) #define DBG(v) cerr << #v << " = " << (v) << endl; #define FOREACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++) #define fi first #define se second const int N = 100100; int n,k; int a[N], b[N]; int m; pii vals[N]; bool flipped; char ans[N]; int t[N]; ll getsolution() { ll ret = 0; ll sum = 0, bestpref = 0; FOR(i,n) { sum += t[i]; ret = max(ret, sum-bestpref); bestpref = min(bestpref, sum); } return ret; } const ll infll = 1000100100LL * 1000100100LL; const int NN = 5010; ll dp[NN][NN]; int prv[NN][NN]; bool ok(ll solution, bool update_ans = false) { //cerr << "\nchecking " << solution << "\n"; FOR(i,n+1) FOR(j,k+1) dp[i][j] = infll; dp[0][0] = 0; FOR(j,k+1) FOR(i,n) { ll h = dp[i][j]; if (h+a[i] <= solution && h+a[i] < dp[i+1][j]) { dp[i+1][j] = max(0LL,h+a[i]); prv[i+1][j] = 0; } if (b[i]>0 && h+a[i]+b[i] <= solution && h+a[i]+b[i] < dp[i+1][j+1]) { dp[i+1][j+1] = max(0LL,h+a[i]+b[i]); prv[i+1][j+1] = 1; } } if (update_ans) { int nn = n, kk = k; while (nn > 0) { int pp = prv[nn][kk]; if (pp) ans[nn-1] = 'A'; nn--; kk-=pp; } } //if (solution == 379179) { // FOR(i,n+1) { // //printf("%d ", i); // FOR(j,k+1) printf("%lld ", (dp[i][j]==infll ? solution+3 : dp[i][j])); // printf("\n"); // } //} //cerr << "there are " << SZ(dp[n][0]) << " elements so solution = " << solution << " is " << (dp[n][0].empty() ? "not " : "") << "ok\n"; return dp[n][k] < infll; } int main() { scanf("%d%d", &n, &k); FOR(i,n) scanf("%d", &a[i]); FOR(i,n) scanf("%d", &b[i]); int cA = 0, cB = 0; FOR(i,n) { cA += a[i] < b[i]; cB += a[i] > b[i]; t[i] = min(a[i], b[i]); } int cS = n-cA-cB; ll minsolution = getsolution(); /// easy case: taking min is solution if (cA <= k && k <= cA+cS) { cS = k-cA; FOR(i,n) { if (a[i] < b[i]) ans[i] = 'A'; else if (a[i] > b[i]) ans[i] = 'B'; else { if (cS > 0) ans[i] = 'A'; else ans[i] = 'B'; cS--; } } printf("%lld\n%s\n", minsolution, ans); return 0; } /// flip solution if (cA > k) { flipped = 1; FOR(i,n) swap(a[i], b[i]); swap(cA,cB); k = n-k; } /// reduce FOR(i,n) { if (a[i] <= b[i]) { ans[i] = 'A'; b[i] = 0; } else { ans[i] = 'B'; swap(a[i], b[i]); b[i] = b[i]-a[i]; vals[m++] = mp(b[i], i); } } k -= cA + cS; sort(vals, vals+m); /// worst case FOR(i,n) t[i] = a[i]+b[i]; ll maxsolution = getsolution(); if (maxsolution <= 0) { FOR(i,n) if (k>0 && b[i]>0) { ans[i] = 'A'; a[i] += b[i]; k--; } if (flipped) FOR(i,n) { if (ans[i] == 'A') ans[i] = 'B'; else ans[i] = 'A'; } printf("0\n%s\n", ans); return 0; } //cerr << "new problem is: flip at least " << k << " with\nbase = "; //FOR(i,n) cerr << a[i] << " "; //cerr << "\nadd = "; //FOR(i,n) cerr << b[i] << " "; //cerr << "\n"; //cerr << "solution is in [" << minsolution << ", " << maxsolution << "]\n"; /// find score minsolution--; while (minsolution + 1 < maxsolution) { ll midsolution = (minsolution + maxsolution) / 2; if (ok(midsolution)) { maxsolution = midsolution; } else { minsolution = midsolution; } } /// find answer ok(maxsolution, true); /// flip and print if (flipped) FOR(i,n) { if (ans[i] == 'A') ans[i] = 'B'; else ans[i] = 'A'; } printf("%lld\n%s\n", maxsolution, ans); return 0; }
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 177 178 179 | #include <bits/stdc++.h> using namespace std; #define REP(i,a,b) for (int i = (a); i <= (b); ++i) #define REPD(i,a,b) for (int i = (a); i >= (b); --i) #define FORI(i,n) REP(i,1,n) #define FOR(i,n) REP(i,0,int(n)-1) #define mp make_pair #define pb push_back #define pii pair<int,int> #define vi vector<int> #define ll long long #define SZ(x) int((x).size()) #define DBG(v) cerr << #v << " = " << (v) << endl; #define FOREACH(i,t) for (typeof(t.begin()) i=t.begin(); i!=t.end(); i++) #define fi first #define se second const int N = 100100; int n,k; int a[N], b[N]; int m; pii vals[N]; bool flipped; char ans[N]; int t[N]; ll getsolution() { ll ret = 0; ll sum = 0, bestpref = 0; FOR(i,n) { sum += t[i]; ret = max(ret, sum-bestpref); bestpref = min(bestpref, sum); } return ret; } const ll infll = 1000100100LL * 1000100100LL; const int NN = 5010; ll dp[NN][NN]; int prv[NN][NN]; bool ok(ll solution, bool update_ans = false) { //cerr << "\nchecking " << solution << "\n"; FOR(i,n+1) FOR(j,k+1) dp[i][j] = infll; dp[0][0] = 0; FOR(j,k+1) FOR(i,n) { ll h = dp[i][j]; if (h+a[i] <= solution && h+a[i] < dp[i+1][j]) { dp[i+1][j] = max(0LL,h+a[i]); prv[i+1][j] = 0; } if (b[i]>0 && h+a[i]+b[i] <= solution && h+a[i]+b[i] < dp[i+1][j+1]) { dp[i+1][j+1] = max(0LL,h+a[i]+b[i]); prv[i+1][j+1] = 1; } } if (update_ans) { int nn = n, kk = k; while (nn > 0) { int pp = prv[nn][kk]; if (pp) ans[nn-1] = 'A'; nn--; kk-=pp; } } //if (solution == 379179) { // FOR(i,n+1) { // //printf("%d ", i); // FOR(j,k+1) printf("%lld ", (dp[i][j]==infll ? solution+3 : dp[i][j])); // printf("\n"); // } //} //cerr << "there are " << SZ(dp[n][0]) << " elements so solution = " << solution << " is " << (dp[n][0].empty() ? "not " : "") << "ok\n"; return dp[n][k] < infll; } int main() { scanf("%d%d", &n, &k); FOR(i,n) scanf("%d", &a[i]); FOR(i,n) scanf("%d", &b[i]); int cA = 0, cB = 0; FOR(i,n) { cA += a[i] < b[i]; cB += a[i] > b[i]; t[i] = min(a[i], b[i]); } int cS = n-cA-cB; ll minsolution = getsolution(); /// easy case: taking min is solution if (cA <= k && k <= cA+cS) { cS = k-cA; FOR(i,n) { if (a[i] < b[i]) ans[i] = 'A'; else if (a[i] > b[i]) ans[i] = 'B'; else { if (cS > 0) ans[i] = 'A'; else ans[i] = 'B'; cS--; } } printf("%lld\n%s\n", minsolution, ans); return 0; } /// flip solution if (cA > k) { flipped = 1; FOR(i,n) swap(a[i], b[i]); swap(cA,cB); k = n-k; } /// reduce FOR(i,n) { if (a[i] <= b[i]) { ans[i] = 'A'; b[i] = 0; } else { ans[i] = 'B'; swap(a[i], b[i]); b[i] = b[i]-a[i]; vals[m++] = mp(b[i], i); } } k -= cA + cS; sort(vals, vals+m); /// worst case FOR(i,n) t[i] = a[i]+b[i]; ll maxsolution = getsolution(); if (maxsolution <= 0) { FOR(i,n) if (k>0 && b[i]>0) { ans[i] = 'A'; a[i] += b[i]; k--; } if (flipped) FOR(i,n) { if (ans[i] == 'A') ans[i] = 'B'; else ans[i] = 'A'; } printf("0\n%s\n", ans); return 0; } //cerr << "new problem is: flip at least " << k << " with\nbase = "; //FOR(i,n) cerr << a[i] << " "; //cerr << "\nadd = "; //FOR(i,n) cerr << b[i] << " "; //cerr << "\n"; //cerr << "solution is in [" << minsolution << ", " << maxsolution << "]\n"; /// find score minsolution--; while (minsolution + 1 < maxsolution) { ll midsolution = (minsolution + maxsolution) / 2; if (ok(midsolution)) { maxsolution = midsolution; } else { minsolution = midsolution; } } /// find answer ok(maxsolution, true); /// flip and print if (flipped) FOR(i,n) { if (ans[i] == 'A') ans[i] = 'B'; else ans[i] = 'A'; } printf("%lld\n%s\n", maxsolution, ans); return 0; } |