#include <cstdio> #include <cassert> #include <cinttypes> #include <set> #define MAXN 100200 typedef long long ll; ll inputA[MAXN]; ll inputB[MAXN]; int N; int K; ll *A; ll *B; char Achar, Bchar; char output[MAXN]; ll Add(ll x, ll y) { return (x + y > 0 ? x + y : 0); } typedef struct movement { ll val; int pos; } movement; bool operator< (const movement &A, const movement &B) { return (A.val < B.val) || (A.val == B.val && A.pos < B.pos); } bool CanAchieve(ll maxval) { ll optimalplay = 0; ll lastinset = 0; std::set<movement> current; int rejected = 0; for (int i = 0; i < N; ++i) { int newA = A[i]; int newB = B[i]; if (newB < newA) newA = newB; while (!current.empty() && Add(optimalplay + current.begin()->val, newA) == Add(optimalplay, newA)) { // Check if the first entry flattens by horizontal movement. assert(Add(optimalplay + current.begin()->val, newA) == 0); optimalplay += current.begin()->val; current.erase(current.begin()); if (current.empty()) { assert(optimalplay == lastinset); } } // OK, now horizontals are cleared. Let's check if the vertical disappears. if (Add(optimalplay, newA) == Add(optimalplay, newB)) { // OK, our thing disappears into the flatten as well. Toss it, we'll take newB for everything. if (!current.empty()) { movement first = *current.begin(); current.erase(current.begin()); ll newval = Add(optimalplay + first.val, newB) - Add(optimalplay, newB); assert(newval > 0LL); assert(newval <= first.val); first.val = newval; current.insert(first); assert(current.begin()->pos == first.pos); } } else { // We will be extending the current set by our value. We just need to check whether it will be the // new first value (in which case we need to mangle it's val) or not (in which case we mangle the first's val). movement newmovement; newmovement.pos = i; if (current.empty()) { // OK, this is the easy case, we select our new thing :) newmovement.val = Add(optimalplay, newB) - Add(optimalplay, newA); assert(newmovement.val > 0LL); current.insert(newmovement); } else { if (Add(optimalplay, newB) > Add(optimalplay + current.begin()->val, newA)) { // In this case, the vertical movement is worse than the horizontal. newmovement.val = newB - newA; assert(newmovement.val >= current.begin()->val); current.insert(newmovement); movement updatedfirst = *(current.begin()); assert(updatedfirst.pos < i); current.erase(current.begin()); ll newval = Add(optimalplay + updatedfirst.val, newA) - Add(optimalplay, newA); assert(newval > 0LL); assert(newval <= updatedfirst.val); assert(newval <= newmovement.val); updatedfirst.val = newval; current.insert(updatedfirst); } else { // In this case, we choose the vertical movement - so no change to the current first, just need to mangle // the new first correctly. newmovement.val = Add(optimalplay, newB) - Add(optimalplay, newA); assert(newmovement.val > 0LL); assert(newmovement.val <= newB - newA); assert(newmovement.val <= current.begin()->val); current.insert(newmovement); } } } optimalplay = Add(optimalplay, newA); lastinset = Add(lastinset, newB); while (lastinset > maxval) { if (current.empty()) { assert(optimalplay == lastinset); return false; } movement ditched = *(current.rbegin()); current.erase(--current.end()); // You can't erase a reverse_iterator, which is annoying. lastinset -= ditched.val; output[ditched.pos] = Achar; rejected += 1; } if (rejected > K) return false; } return true; } int main() { scanf("%d %d", &N, &K); for (int i = 0; i < N; ++i) scanf("%" PRId64, &inputA[i]); for (int i = 0; i < N; ++i) scanf("%" PRId64, &inputB[i]); int inputAbetter = 0; for (int i = 0; i < N; ++i) if (inputA[i] < inputB[i]) inputAbetter++; // A are the ones I want to take, B are the ones I have to take (N-K) of (I can drop at most K). A = inputA; B = inputB; Achar = 'A'; Bchar = 'B'; if (inputAbetter < K) { B = inputA; A = inputB; K = N - K; Achar = 'B'; Bchar = 'A'; } // We binary search to find the lowest value we can limit the thieves to. // We know we can limit them to N * 10**9 (trivially, by doing whatever). // We also know we can't limit them to -1. // At every point, lo can't be achieved, while hi can. ll lo = -1; ll hi = N * 1000000000LL; while (hi - lo > 1) { ll med = (hi + lo) / 2; if (CanAchieve(med)) { hi = med; } else { lo = med; } } int Bcount = 0; for (int i = 0; i < N; ++i) { output[i] = 'C'; if (B[i] < A[i]) { output[i] = Bchar; Bcount += 1; } } assert(CanAchieve(hi)); // Just to fill in the Achars - the ones we dropped. for (int i = 0; i < N; ++i) { if (output[i] == 'C' && Bcount < (N-K)) { output[i] = Bchar; Bcount += 1; } else { if (output[i] == 'C') output[i] = Achar; } } // printf("%d %d %d\n", N, K, Bcount); assert(Bcount == N - K); output[N] = 0; printf("%" PRId64 "\n%s\n", hi, output); 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 | #include <cstdio> #include <cassert> #include <cinttypes> #include <set> #define MAXN 100200 typedef long long ll; ll inputA[MAXN]; ll inputB[MAXN]; int N; int K; ll *A; ll *B; char Achar, Bchar; char output[MAXN]; ll Add(ll x, ll y) { return (x + y > 0 ? x + y : 0); } typedef struct movement { ll val; int pos; } movement; bool operator< (const movement &A, const movement &B) { return (A.val < B.val) || (A.val == B.val && A.pos < B.pos); } bool CanAchieve(ll maxval) { ll optimalplay = 0; ll lastinset = 0; std::set<movement> current; int rejected = 0; for (int i = 0; i < N; ++i) { int newA = A[i]; int newB = B[i]; if (newB < newA) newA = newB; while (!current.empty() && Add(optimalplay + current.begin()->val, newA) == Add(optimalplay, newA)) { // Check if the first entry flattens by horizontal movement. assert(Add(optimalplay + current.begin()->val, newA) == 0); optimalplay += current.begin()->val; current.erase(current.begin()); if (current.empty()) { assert(optimalplay == lastinset); } } // OK, now horizontals are cleared. Let's check if the vertical disappears. if (Add(optimalplay, newA) == Add(optimalplay, newB)) { // OK, our thing disappears into the flatten as well. Toss it, we'll take newB for everything. if (!current.empty()) { movement first = *current.begin(); current.erase(current.begin()); ll newval = Add(optimalplay + first.val, newB) - Add(optimalplay, newB); assert(newval > 0LL); assert(newval <= first.val); first.val = newval; current.insert(first); assert(current.begin()->pos == first.pos); } } else { // We will be extending the current set by our value. We just need to check whether it will be the // new first value (in which case we need to mangle it's val) or not (in which case we mangle the first's val). movement newmovement; newmovement.pos = i; if (current.empty()) { // OK, this is the easy case, we select our new thing :) newmovement.val = Add(optimalplay, newB) - Add(optimalplay, newA); assert(newmovement.val > 0LL); current.insert(newmovement); } else { if (Add(optimalplay, newB) > Add(optimalplay + current.begin()->val, newA)) { // In this case, the vertical movement is worse than the horizontal. newmovement.val = newB - newA; assert(newmovement.val >= current.begin()->val); current.insert(newmovement); movement updatedfirst = *(current.begin()); assert(updatedfirst.pos < i); current.erase(current.begin()); ll newval = Add(optimalplay + updatedfirst.val, newA) - Add(optimalplay, newA); assert(newval > 0LL); assert(newval <= updatedfirst.val); assert(newval <= newmovement.val); updatedfirst.val = newval; current.insert(updatedfirst); } else { // In this case, we choose the vertical movement - so no change to the current first, just need to mangle // the new first correctly. newmovement.val = Add(optimalplay, newB) - Add(optimalplay, newA); assert(newmovement.val > 0LL); assert(newmovement.val <= newB - newA); assert(newmovement.val <= current.begin()->val); current.insert(newmovement); } } } optimalplay = Add(optimalplay, newA); lastinset = Add(lastinset, newB); while (lastinset > maxval) { if (current.empty()) { assert(optimalplay == lastinset); return false; } movement ditched = *(current.rbegin()); current.erase(--current.end()); // You can't erase a reverse_iterator, which is annoying. lastinset -= ditched.val; output[ditched.pos] = Achar; rejected += 1; } if (rejected > K) return false; } return true; } int main() { scanf("%d %d", &N, &K); for (int i = 0; i < N; ++i) scanf("%" PRId64, &inputA[i]); for (int i = 0; i < N; ++i) scanf("%" PRId64, &inputB[i]); int inputAbetter = 0; for (int i = 0; i < N; ++i) if (inputA[i] < inputB[i]) inputAbetter++; // A are the ones I want to take, B are the ones I have to take (N-K) of (I can drop at most K). A = inputA; B = inputB; Achar = 'A'; Bchar = 'B'; if (inputAbetter < K) { B = inputA; A = inputB; K = N - K; Achar = 'B'; Bchar = 'A'; } // We binary search to find the lowest value we can limit the thieves to. // We know we can limit them to N * 10**9 (trivially, by doing whatever). // We also know we can't limit them to -1. // At every point, lo can't be achieved, while hi can. ll lo = -1; ll hi = N * 1000000000LL; while (hi - lo > 1) { ll med = (hi + lo) / 2; if (CanAchieve(med)) { hi = med; } else { lo = med; } } int Bcount = 0; for (int i = 0; i < N; ++i) { output[i] = 'C'; if (B[i] < A[i]) { output[i] = Bchar; Bcount += 1; } } assert(CanAchieve(hi)); // Just to fill in the Achars - the ones we dropped. for (int i = 0; i < N; ++i) { if (output[i] == 'C' && Bcount < (N-K)) { output[i] = Bchar; Bcount += 1; } else { if (output[i] == 'C') output[i] = Achar; } } // printf("%d %d %d\n", N, K, Bcount); assert(Bcount == N - K); output[N] = 0; printf("%" PRId64 "\n%s\n", hi, output); return 0; } |