The 18th Zhejiang University Programming Contest Sponsored by TuSimple
A. Pretty Matrix
题意：模拟
方法：
code：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <iomanip> 5 6 #include <vector> 7 #include <cstring> 8 #include <string> 9 #include <queue> 10 #include <deque> 11 #include <stack> 12 #include <map> 13 #include <set> 14 #include <unordered_map> 15 #include <unordered_set> 16 #include <utility> 17 18 #include <cmath> 19 #include <algorithm> 20 #include <cassert> 21 #include <bitset> 22 using namespace std; 23 24 typedef long long ll; 25 typedef pair<int, int> ii; 26 typedef pair<ll, ll> l4; 27 28 #define mp make_pair 29 #define pb push_back 30 31 32 const int maxn = 1e2; 33 int a[maxn][maxn], n, m, A, B; 34 35 int main() 36 { 37 int T; 38 scanf("%d", &T); 39 for (int kase = 1; kase <= T; ++kase) 40 { 41 scanf("%d %d %d %d", &n, &m, &A, &B); 42 int ans = 0; 43 for (int i = 0; i < n; ++i) 44 for (int j = 0; j < m; ++j) 45 { 46 scanf("%d", &a[i][j]); 47 ans += a[i][j] < A || a[i][j] > B; 48 } 49 if (A > B) 50 { 51 puts("No Solution"); 52 } 53 else 54 printf("%d\n", ans); 55 } 56 }
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B. Liblume
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C. Mergeable Stack
题意：n个stack，q个操作。（n, q <= 3e5)。操作有三种，stack[i].push(v); stack[i].pop(); insert stack[i] to stack[j]。
方法：链表。
code：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <iomanip> 5 6 #include <vector> 7 #include <cstring> 8 #include <string> 9 #include <queue> 10 #include <deque> 11 #include <stack> 12 #include <map> 13 #include <set> 14 #include <unordered_map> 15 #include <unordered_set> 16 #include <utility> 17 #include <list> 18 19 #include <cmath> 20 #include <algorithm> 21 #include <cassert> 22 #include <bitset> 23 using namespace std; 24 25 typedef long long ll; 26 typedef pair<int, int> ii; 27 typedef pair<ll, ll> l4; 28 29 #define mp make_pair 30 #define pb push_back 31 32 const int maxn = 3e5+1; 33 34 list<int> s[maxn]; 35 int n, q; 36 const int PUSH = 1, POP = 2, MOVE = 3; 37 int main() 38 { 39 int T; 40 scanf("%d", &T); 41 for (int kase = 1; kase <= T; ++kase) 42 { 43 scanf("%d %d", &n, &q); 44 for (int i = 1; i <= n; ++i) 45 s[i].clear(); 46 for (int i = 0; i < q; ++i) 47 { 48 int op; 49 scanf("%d", &op); 50 if (op == PUSH) 51 { 52 int id, v; 53 scanf("%d %d", &id, &v); 54 s[id].push_back(v); 55 } 56 else if (op == POP) 57 { 58 int id; 59 scanf("%d", &id); 60 if (s[id].empty()) 61 puts("EMPTY"); 62 else 63 { 64 int v = s[id].back(); 65 s[id].pop_back(); 66 printf("%d\n", v); 67 } 68 } 69 else if (op == MOVE) 70 { 71 int dest, src; 72 scanf("%d %d", &dest, &src); 73 s[dest].splice(s[dest].end(), s[src]); 74 } 75 else 76 assert(false); 77 } 78 } 79 }
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D. RAID-ZOJ
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E. Crosses Puzzles
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F. Schrodinger's Puzzles
题意：一种特殊的背包，容量为c。每个物品有一个大小sz和系数k，放入背包时价值是k*剩余容量。有两种物品，第一种物品的k都是k1， 第二种都是k2。问你背包最大价值是多少。
方法：首先现将同种的物品按照sz从小到大排序。枚举第一种物品放了几个，这样第二个物品放了几个也可以确定。下面就是将两种物品的清单merge到一起。
code：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <iomanip> 5 6 #include <vector> 7 #include <cstring> 8 #include <string> 9 #include <queue> 10 #include <deque> 11 #include <stack> 12 #include <map> 13 #include <set> 14 #include <unordered_map> 15 #include <unordered_set> 16 #include <utility> 17 #include <list> 18 19 #include <cmath> 20 #include <algorithm> 21 #include <cassert> 22 #include <bitset> 23 using namespace std; 24 25 typedef long long ll; 26 typedef pair<int, int> ii; 27 typedef pair<ll, ll> l4; 28 29 #define mp make_pair 30 #define pb push_back 31 32 const int maxn = 2e3; 33 int s[2][maxn], k[2], n[2], c; 34 int tots[maxn<<1]; 35 int main() 36 { 37 int T; 38 scanf("%d", &T); 39 for (int kase = 1; kase <= T; ++kase) 40 { 41 for (int i = 0; i < 2; ++i) 42 scanf("%d", k+i); 43 scanf("%d", &c); 44 for (int i = 0; i < 2; ++i) 45 scanf("%d", n+i); 46 for (int i = 0; i < 2; ++i) 47 { 48 for (int j = 0; j < n[i]; ++j) 49 scanf("%d", &s[i][j]); 50 sort(s[i], s[i] + n[i]); 51 } 52 ll ans = 0; 53 for (int sz0 = 0; sz0 <= n[0]; ++sz0) 54 { 55 ll sz = 0; 56 for (int i = 0; i < sz0; ++i) 57 sz += s[0][i]; 58 if (sz > c) 59 break; 60 int sz1 = 0; 61 while (sz1 < n[1] && sz + s[1][sz1] <= c) 62 { 63 sz += s[1][sz1]; 64 ++sz1; 65 } 66 ll tmp = 0; 67 int p0 = 0, p1 = 0; 68 sz = c; 69 while (p0 < sz0 && p1 < sz1) 70 { 71 if (1ll * k[0] * s[1][p1] > 1ll * k[1] * s[0][p0]) 72 { 73 //pick p0 74 sz -= s[0][p0++]; 75 tmp += sz * k[0]; 76 } 77 else 78 { 79 sz -= s[1][p1++]; 80 tmp += sz * k[1]; 81 } 82 } 83 while (p0 < sz0) 84 { 85 sz -= s[0][p0++]; 86 tmp += sz * k[0]; 87 } 88 while (p1 < sz1) 89 { 90 sz -= s[1][p1++]; 91 tmp += sz * k[1]; 92 } 93 if (tmp > ans) 94 ans = tmp; 95 } 96 printf("%lld\n", ans); 97 } 98 }
WA：也可以用dp[i][j]表示第一种物品选了前i个，第二种选了前j个后，得到的最大价值。比较好写。dp[i][j] 只会由dp[i-1][j]和dp[i][j-1]更新过来。
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <iomanip> 5 6 #include <vector> 7 #include <cstring> 8 #include <string> 9 #include <queue> 10 #include <deque> 11 #include <stack> 12 #include <map> 13 #include <set> 14 #include <unordered_map> 15 #include <unordered_set> 16 #include <utility> 17 #include <list> 18 19 #include <cmath> 20 #include <algorithm> 21 #include <cassert> 22 #include <bitset> 23 using namespace std; 24 25 typedef long long ll; 26 typedef pair<int, int> ii; 27 typedef pair<ll, ll> l4; 28 29 #define mp make_pair 30 #define pb push_back 31 32 const int maxn = 2e3+1; 33 ll s[2][maxn], k[2], n[2], c, sum[2][maxn]; 34 ll d[maxn][maxn]; 35 inline void Max(ll &a, ll b) 36 { 37 if (a < b) a = b; 38 } 39 int main() 40 { 41 int T; 42 scanf("%d", &T); 43 for (int kase = 1; kase <= T; ++kase) 44 { 45 for (int i = 0; i < 2; ++i) 46 scanf("%lld", k+i); 47 scanf("%lld", &c); 48 for (int i = 0; i < 2; ++i) 49 scanf("%lld", n+i); 50 for (int i = 0; i < 2; ++i) 51 { 52 for (int j = 1; j <= n[i]; ++j) 53 { 54 scanf("%lld", &s[i][j]); 55 } 56 sort(s[i]+1, s[i]+1+n[i]); 57 for (int j = 1; j <= n[i]; ++j) 58 sum[i][j] = sum[i][j-1] + s[i][j]; 59 } 60 ll ans = 0; 61 for (int i = 0; i <= n[0]; ++i) 62 for (int j = 0; j <= n[1]; ++j) 63 d[i][j] = 0; 64 for (int i = 0; i <= n[0]; ++i) 65 for (int j = 0; j <= n[1]; ++j) 66 { 67 ll curc = sum[0][i] + sum[1][j]; 68 if (curc <= c) 69 { 70 Max(ans, d[i][j]); 71 if (i < n[0]) 72 Max(d[i+1][j], d[i][j] + (c - curc - s[0][i+1]) * k[0]); 73 if (j < n[1]) 74 Max(d[i][j+1], d[i][j] + (c - curc - s[1][j+1]) * k[1]); 75 } 76 } 77 printf("%lld\n", ans); 78 } 79 }
G. Traffic Light
题意：
方法：bfs
code：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <iomanip> 5 6 #include <vector> 7 #include <cstring> 8 #include <string> 9 #include <queue> 10 #include <deque> 11 #include <stack> 12 #include <map> 13 #include <set> 14 #include <unordered_map> 15 #include <unordered_set> 16 #include <utility> 17 #include <list> 18 19 #include <cmath> 20 #include <algorithm> 21 #include <cassert> 22 #include <bitset> 23 using namespace std; 24 25 typedef long long ll; 26 typedef pair<int, int> ii; 27 typedef pair<ll, ll> l4; 28 29 #define mp make_pair 30 #define pb push_back 31 32 const int maxn = 1e5; 33 int d[maxn][2], n, m, t[maxn], tot; 34 ii src, dest; 35 inline int id(const ii &pr) 36 { 37 return pr.first * m + pr.second; 38 } 39 int dx[2][2] = 40 { 41 {-1, 1}, {0, 0} 42 }; 43 int dy[2][2] = 44 { 45 {0, 0}, {-1, 1} 46 }; 47 48 queue<pair<ii, int> > q; 49 inline void touch(ii &pr, int dist) 50 { 51 if (d[id(pr)][dist&1] == -1) 52 { 53 d[id(pr)][dist&1] = dist; 54 q.push(mp(pr, dist)); 55 } 56 57 } 58 int main() 59 { 60 int T; 61 scanf("%d", &T); 62 for (int kase = 1; kase <= T; ++kase) 63 { 64 scanf("%d %d", &n, &m); 65 tot = n * m; 66 for (int i = 0; i < tot; ++i) 67 { 68 scanf("%d", t+i); 69 d[i][0] = d[i][1] = -1; 70 } 71 scanf("%d %d %d %d", &src.first, &src.second, &dest.first, &dest.second); 72 --src.first, --src.second, --dest.first, --dest.second; 73 assert(q.empty()); 74 touch(src, 0); 75 while (!q.empty()) 76 { 77 ii cur = q.front().first; 78 int dist = q.front().second; 79 q.pop(); 80 int curt = (t[id(cur)] ^ dist) & 1; 81 int nxtdist = dist + 1; 82 for (int i = 0; i < 2; ++i) 83 { 84 ii nxt; 85 nxt.first = cur.first + dx[curt][i]; 86 nxt.second = cur.second + dy[curt][i]; 87 if (nxt.first < 0 || nxt.second < 0 || nxt.first >= n || nxt.second >= m) 88 continue; 89 touch(nxt, nxtdist); 90 } 91 } 92 int ans = d[id(dest)][0]; 93 int ans2 = d[id(dest)][1]; 94 if (ans2 != -1) 95 if (ans == -1 || ans > ans2) 96 ans = ans2; 97 printf("%d\n", ans); 98 } 99 }
WA：注意处理输入中1-based的坐标。记得检查坐标是否合法。
H. Boolean Expression
题意：
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I. Honorifics
题意：
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J. PPAP
题意：模拟
方法：
code：
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 #include <iomanip> 5 6 #include <vector> 7 #include <cstring> 8 #include <string> 9 #include <queue> 10 #include <deque> 11 #include <stack> 12 #include <map> 13 #include <set> 14 #include <unordered_map> 15 #include <unordered_set> 16 #include <utility> 17 18 #include <cmath> 19 #include <algorithm> 20 #include <cassert> 21 #include <bitset> 22 using namespace std; 23 24 typedef long long ll; 25 typedef pair<int, int> ii; 26 typedef pair<ll, ll> l4; 27 28 #define mp make_pair 29 #define pb push_back 30 31 int main() 32 { 33 ios::sync_with_stdio(false); 34 cin.tie(0); 35 int T; 36 cin >> T; 37 for (int kase = 1; kase <= T; ++kase) 38 { 39 string a, b; 40 cin >> a >> b; 41 b[0] = toupper(b[0]); 42 cout << b << a << '\n'; 43 } 44 }
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