题面
预计得分：70 + 60 + 30 = 160
实际得分：40 + 60 + 0 = 100
T1数组开小了
T2比赛结束后5min AC
T3加了个记忆话搜索wa了、、
T1
zbq吊打std啊Orz
此题$O(nlog)$做法：
一个很显然的思路：对每个做括号维护一个大根堆，每次取最大的。
但是这样有不优的情况，比如$()), 1, 3, 5$
那么我们还需要对每个已经加入的右括号维护一个小根堆。每次判断是否替换掉更小的会更优
#include<cstdio> #include<algorithm> #include<queue> #include<vector> #define LL long long using namespace std; const int MAXN = 2 * 1e5 + 10, INF = 1e9; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int N; priority_queue<int> mx; priority_queue<int, vector<int>, greater<int> >mi; char s[MAXN]; int a[MAXN]; int main() { freopen("bracket.in", "r", stdin); freopen("bracket.out", "w", stdout); N = read(); scanf("%s", s + 1); for(int i = 1; i <= N; i++) a[i] = read(); int ans = 0; for(int i = 1; i <= N; i++) { if(s[i] == '(') mx.push(a[i]); if(s[i] == ')') if(!mx.empty() && a[i] + mx.top() > 0) { if(mi.empty() || (!mi.empty() && mx.top() > - mi.top())) ans += a[i] + mx.top(), mx.pop(), mi.push(a[i]); else if(!mi.empty() && a[i] > mi.top()) ans -= mi.top(), mi.pop(), ans += a[i], mi.push(a[i]); } else if(!mi.empty() && a[i] > mi.top()) ans -= mi.top(), mi.pop(), ans += a[i], mi.push(a[i]); } printf("%d", ans); return 0; }
T2
很显然每个位置就那么几种可能
直接暴力判断就好，前缀和优化
/* 60：直接BFS */ #include<cstdio> #include<algorithm> #include<queue> #define LL long long using namespace std; const int MAXN = 1e5 + 10, INF = 1e9; inline int read() { char c = getchar(); int x = 0, f = 1; while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();} while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar(); return x * f; } int N, M, K; int a[1001][1001], down[MAXN], vis[1001][1001]; struct Node { int xx1, yy1, xx2, yy2; }p[MAXN]; bool pd(int x, int y) { if(x < 1 || x > N || y < 1 || y > M) return 1; return 0; } int hsum[1001][1001], lsum[1001][1001]; bool line(int x1, int y11, int x2, int y2, int id) { if(pd(x1, y11) || pd(x2, y2)) return 1; if(x1 == x2) { if(y11 > y2) swap(y11, y2); if(hsum[x1][y2] - hsum[x1][y11 - 1] == 0) return 0; if(a[x1][y11] == id && hsum[x1][y2] - hsum[x1][y11] == 0) return 0; if(a[x1][y2] == id && hsum[x1][y2 - 1] - hsum[x1][y11 - 1] == 0) return 0; return 1; } if(y11 == y2) { if(x1 > x2) swap(x1, x2); if(lsum[x2][y2] - lsum[x1 - 1][y2] == 0) return 0; if(a[x1][y11] == id && lsum[x2][y2] - lsum[x1][y2] == 0) return 0; if(a[x2][y11] == id && lsum[x2 - 1][y2] - lsum[x1 - 1][y2] ==0) return 0; return 1; } } int main() { freopen("linking.in", "r", stdin); freopen("linking.out", "w", stdout); N = read(); M = read(); K = read(); for(int i = 1; i <= K; i++) { int xx1 = read(), yy1 = read(), xx2 = read(), yy2 = read(); a[xx1][yy1] = i; a[xx2][yy2] = i; p[i] = (Node) {xx1, yy1, xx2, yy2}; } for(int i = 1; i <= N; i++) for(int j = 1; j <= M; j++) hsum[i][j] = hsum[i][j - 1] + a[i][j], lsum[i][j] = lsum[i - 1][j] + a[i][j]; int ans = 0; for(int i = 1; i <= K; i++) { int xx1 = p[i].xx1, yy1 = p[i].yy1, xx2 = p[i].xx2, yy2 = p[i].yy2, flag = 0; if(yy1 > yy2) swap(yy1, yy2), swap(xx1, xx2); for(int k = 1; k <= N; k++) if(!line(xx2, yy2, k, yy2, i) && !line(xx1, yy1, k, yy1, i) && !line(k, yy1, k, yy2, i)) {ans++; flag = 1; break;} if(flag == 1) continue; for(int k = 1; k <= M; k++) if(!line(xx2, yy2, xx2, k, i) && !line(xx2, k, xx1, k, i) && !line(xx1, yy1, xx1, k, i)) {ans++; break;} } printf("%d", ans); return 0; } /* 20 20 3 1 1 20 20 2 1 2 20 3 1 1 20 3 3 3 1 3 2 2 1 1 3 3 1 2 2 1 */
T3
神仙题。
很显然答案是一棵树，那么直接书上倍增就好
满分做法不会。。