省赛选拔，打之前觉得好神仙，打之中觉得这些题真tm都不错，打完之后看题解为什么这些傻逼题都没出，QAQ
记录一些值得学习的自我感觉有意义补的题
Day1。血崩3题
1001 搬东西 赛后补题（血亏）
给你2N+K个物品，可以选K个放在车上，剩下2N个自己搬，每次搬两个，消耗是两个物品的差的绝对值，求最小消耗
标程：DP。我：贪心+DP，一个（假）结论，应该之前是做过原题的
设dp（i，j）表示前i个物品，j个放在车上的最小消耗，答案是dp（2n+k，k），决策都是放车和不放车
标程给的转移是选相邻两个物品或者间隔一个的物品，不过好像我猜了一下最优解一定是相邻两个一起搬
所以dp方程就变成了：dp[i][j] = min(dp[i-1][j-1]/*表示放*/, dp[i-2][j] + (a[i]-a[i-1])/*表示搬*/)
可能是水过去的AC代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #define INF 0x3f3f3f3f 6 #define LL long long 7 #define debug(x) cout << "[" << x << "]" << endl 8 using namespace std; 9 10 const int mx = 1e5+25; 11 LL dp[2*mx][25]; 12 int a[2*mx]; 13 14 int main(){ 15 int n, k; 16 scanf("%d%d", &n, &k); 17 memset(dp, INF, sizeof dp); 18 for (int i = 1; i <= 2*n+k; i++) scanf("%d", &a[i]); 19 sort(a+1, a+n*2+k+1); 20 for (int i = 0; i <= k; i++) dp[i][i] = 0; 21 for (int i = 2; i <= n*2+k; i++){ 22 for (int j = 0; j <= k; j++){ 23 if (j) dp[i][j] = min(dp[i-1][j-1], dp[i][j]); 24 dp[i][j] = min(dp[i][j], dp[i-2][j]+a[i]-a[i-1]); 25 } 26 } 27 printf("%lld ", dp[2*n+k][k]); 28 return 0; 29 }
1003 天气变化 / 1010 气温预测
这两道题居然是一起出的，比赛时放了后面一题，找一个数右边最后一个的比它大的数，前面是找右边第一个比它大的数
1010赛场上是线段树直接先往右子树再往左子树单点查询，1003更好写单调栈一下就行了
1003代码：
咕咕咕
1010比赛提交代码：
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #include<stack> 6 #define lid id << 1 7 #define rid id << 1 | 1 8 #define LL long long 9 #define debug(x) cout << "[" << x << "]" << endl 10 using namespace std; 11 12 const int mx = 1e6+5; 13 struct tree{ 14 int l, r, m; 15 }tree[mx<<2]; 16 int a[mx]; 17 18 void build(int l, int r, int id){ 19 tree[id].l = l; 20 tree[id].r = r; 21 if (l == r){ 22 tree[id].m = a[l]; 23 return; 24 } 25 int mid = (l+r)>>1; 26 build(l, mid, lid); 27 build(mid+1, r, rid); 28 tree[id].m = max(tree[lid].m, tree[rid].m); 29 } 30 31 int query(int id, int x, int i){ 32 if (tree[id].m <= x) return i; 33 if (tree[id].l == tree[id].r) return tree[id].l; 34 if (x < tree[rid].m) return query(rid, x, i); 35 else return query(lid, x, i); 36 } 37 38 int main(){ 39 int n; 40 scanf("%d", &n); 41 for (int i = 1; i <= n; i++) scanf("%d", &a[i]); 42 build (1, n, 1); 43 for (int i = 1; i <= n; i++){ 44 int ans = query(1, a[i], i); 45 if (ans <= i) ans = i; 46 if (i == 1) printf("%d", ans-i); 47 else printf(" %d", ans-i); 48 } 49 printf(" "); 50 return 0; 51 }
1009 模和最大
给一个数列a，对每个a[i]找一个j != i，使得（a[i]+a[j]）%p最大，输出这个值
数据保证a[i] < p
通过小于p可以知道任意两个数的和不超过2p，二分找加起来小于p的最大值，再取一个最大的值相加，取max一定是正解
需要处理一下边界各种细节，赛场上的代码比较乱，也不改了，能出的快就行
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #define LL long long 6 #define debug(x) cout << "[" << x << "]" << endl 7 using namespace std; 8 9 const int mx = 1e6+5; 10 int ans[mx], b[mx]; 11 12 struct node{ 13 int id, num; 14 bool operator < (const node& a) const{ 15 return num < a.num; 16 } 17 }a[mx]; 18 19 int main(){ 20 int n, k; 21 scanf("%d%d", &n, &k); 22 for (int i = 1; i <= n; i++) { 23 scanf("%d", &a[i].num); 24 a[i].id = i; 25 } 26 sort(a+1, a+n+1); 27 for (int i = 1; i <= n; i++) b[i] = a[i].num; 28 for (int i = 1; i <= n; i++){ 29 int s = upper_bound(b+1, b+n+1, k-1-b[i])-b; 30 if (b[s] > k-1-b[i]) s--; 31 if (!s) s = n; 32 if (i == s) s--; 33 if (!s) s = n; 34 ans[a[i].id] = (b[i]+b[s])%k; 35 if (i == n) ans[a[i].id] = max(ans[a[i].id], (b[n]+b[n-1])%k); 36 else ans[a[i].id] = max(ans[a[i].id], (b[i]+b[n])%k); 37 } 38 for (int i = 1; i <= n; i++){ 39 if (i == 1) printf("%d", ans[i]); 40 else printf(" %d", ans[i]); 41 } 42 printf(" "); 43 return 0; 44 }
Day2 5题 比第一次简单好歹还能抢救一下我
赛前还听说这场全都是神仙题，吓死了QAQ
1011 神秘群岛
给一棵树，边是双向的且权值不同，每条边每个方向只能走一次，q次询问u到v能走到的最大边权
n，q 1e5
画个图可以发现除了v到u的简单路径，其它所有的边权都能收集到，跟LCA求链长的区别就是在dfs的时候多算了一条反向边
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<vector> 5 #include<iostream> 6 #define INF 0x3f3f3f3f 7 #define LL long long 8 #define debug(x) cout << "[" << x << "]" << endl 9 using namespace std; 10 11 const int mx = 1e5+5; 12 int d[mx][2], p[mx][30], pa[mx], rk[mx]; 13 int n; 14 15 struct edge{ 16 int v, w, b; 17 edge(int v = 0, int w = 0, int b = 0): v(v), w(w), b(b){} 18 }; 19 vector<edge> G[mx]; 20 21 void dfs(int u, int fa, int cnt){ 22 pa[u] = fa; 23 rk[u] = cnt; 24 int len = G[u].size(); 25 for (int i = 0; i < len; i++){ 26 int v = G[u][i].v; 27 if (v != fa){ 28 d[v][0] = d[u][0]+G[u][i].w; 29 d[v][1] = d[u][1]+G[u][i].b; 30 dfs(v, u, cnt+1); 31 } 32 } 33 } 34 35 void LCA(){ 36 for (int i = 1; i <= n; i++){ 37 p[i][0] = pa[i]; 38 for (int j = 1; (1<<j) <= n; j++) p[i][j] = -1; 39 } 40 for (int j = 1; (1<<j) <= n; j++) 41 for (int i = 1; i <= n; i++) 42 if (p[i][j-1] != -1) p[i][j] = p[p[i][j-1]][j-1]; 43 } 44 45 int query(int x, int y){ 46 if (rk[x] < rk[y]) swap(x, y); 47 int k; 48 for (k = 0; (1<<(k+1)) <= rk[x]; k++); 49 for (int i = k; i >= 0; i--) 50 if (rk[x] - (1<<i) >= rk[y]) x = p[x][i]; 51 if (x == y) return x; 52 for (int i = k; i >= 0; i--){ 53 if (p[x][i] != -1 && p[x][i] != p[y][i]){ 54 x = p[x][i]; 55 y = p[y][i]; 56 } 57 } 58 return p[x][0]; 59 } 60 61 62 int main(){ 63 int t, q, u, v; 64 scanf("%d", &t); 65 while (t--){ 66 int a, b; 67 LL sum = 0; 68 scanf("%d", &n); 69 memset(d, 0, sizeof d); 70 for (int i = 0; i <= n; i++) G[i].clear(); 71 for (int i = 1; i < n; i++){ 72 scanf("%d%d%d%d", &u, &v, &a, &b); 73 G[u].push_back(edge(v, a, b)); 74 G[v].push_back(edge(u, b, a)); 75 sum += (a+b); 76 } 77 dfs(1, -1, 0); 78 LCA(); 79 scanf("%d", &q); 80 while (q--){ 81 scanf("%d%d", &u, &v); 82 int c = query(u, v); 83 printf("%lld ", sum-(d[u][0]+d[v][1]-d[c][1]-d[c][0])); 84 } 85 } 86 return 0; 87 }
1012 笑脸
一个只有：）（的字符串，：）看做笑脸，现在可以将一个前缀翻转一次，求最多能得到多少个笑脸
跑一遍笑脸和反笑脸【（：】的前缀和，枚举断点，要注意边界情况，WA了好几次。。。
1 #include<cstdio> 2 #include<cstring> 3 #include<algorithm> 4 #include<iostream> 5 #define LL long long 6 #define debug(x) cout << "[" << x << "]" << endl 7 using namespace std; 8 9 const int mx = 2e5+10; 10 char s[mx]; 11 int a[mx], b[mx]; 12 13 int main(){ 14 int n; 15 scanf("%d%s", &n, s+1); 16 for (int i = 1; i <= n; i++){ 17 if (s[i-1] == ':' && s[i] == ')') a[i] = a[i-1]+1; 18 else a[i] = a[i-1]; 19 if (s[i-1] == '(' && s[i] == ':') b[i] = b[i-1]+1; 20 else b[i] = b[i-1]; 21 } 22 int ans = a[n]; 23 for (int i = n; i >= 0; i--){ 24 if (s[i] == ':' && s[1] == ':') { 25 ans = max(ans, a[n]-a[i]+b[i]); 26 continue; 27 } 28 else { 29 if (s[i] == ':' && s[i+1] == ')' && s[1] != ':'){ 30 ans = max(ans, a[n]-a[i]+b[i]-1); 31 continue; 32 } 33 if (s[i] != ':' && s[1] == ':' && s[i+1] == ')'){ 34 ans = max(ans, a[n]-a[i]+b[i]+1); 35 continue; 36 } 37 } 38 ans = max(ans, a[n]-a[i]+b[i]); 39 } 40 printf("%d ", ans); 41 return 0; 42 }