#include <cstdio> #include <deque> #include <vector> bool good[2010][2010]; int distance[2010][2010]; std::vector<std::pair<int, int> > STEPS = {{1,0}, {-1,0}, {0,1}, {0, -1}}; int main() { int N,M,K; scanf("%d %d %d\n",&N,&M,&K); for (int i=1; i<=N; ++i) { for (int j=1; j<=M; ++j) { char c; scanf("%c", &c); good[i][j] = (c=='.'); } scanf("\n"); } std::deque<std::pair<int, int> > Q; distance[1][1] = 1; Q.emplace_back(1,1); while (!Q.empty()) { auto pos = Q.front(); Q.pop_front(); for (auto &step : STEPS) { if (distance[pos.first+step.first][pos.second+step.second] == 0 && good[pos.first+step.first][pos.second+step.second]) { distance[pos.first+step.first][pos.second+step.second] = distance[pos.first][pos.second] + 1; Q.emplace_back(pos.first+step.first, pos.second+step.second); } } } int D = distance[N][M]-1; int up = N-1 + M-1; int down = (D-up)/2; up += down; long long int result = 2010*2010*1000000000LL; int cnt = 0; for (int i=0; i<K; ++i) { long long int a,b,cost; scanf("%lld %lld", &a, &b); cost = a*up + b*down; if (cost == result) { cnt++; } else if (cost < result) { result = cost; cnt = 1; } } printf("%lld %d\n", result, cnt); 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 | #include <cstdio> #include <deque> #include <vector> bool good[2010][2010]; int distance[2010][2010]; std::vector<std::pair<int, int> > STEPS = {{1,0}, {-1,0}, {0,1}, {0, -1}}; int main() { int N,M,K; scanf("%d %d %d\n",&N,&M,&K); for (int i=1; i<=N; ++i) { for (int j=1; j<=M; ++j) { char c; scanf("%c", &c); good[i][j] = (c=='.'); } scanf("\n"); } std::deque<std::pair<int, int> > Q; distance[1][1] = 1; Q.emplace_back(1,1); while (!Q.empty()) { auto pos = Q.front(); Q.pop_front(); for (auto &step : STEPS) { if (distance[pos.first+step.first][pos.second+step.second] == 0 && good[pos.first+step.first][pos.second+step.second]) { distance[pos.first+step.first][pos.second+step.second] = distance[pos.first][pos.second] + 1; Q.emplace_back(pos.first+step.first, pos.second+step.second); } } } int D = distance[N][M]-1; int up = N-1 + M-1; int down = (D-up)/2; up += down; long long int result = 2010*2010*1000000000LL; int cnt = 0; for (int i=0; i<K; ++i) { long long int a,b,cost; scanf("%lld %lld", &a, &b); cost = a*up + b*down; if (cost == result) { cnt++; } else if (cost < result) { result = cost; cnt = 1; } } printf("%lld %d\n", result, cnt); return 0; } |