标签:st表 seq const printf codeforce str class long mes
题目大意:给你两个长度为n的数组a, b,问你有多少个问你有多少个区间满足
a中最大值等于b中最小值。
思路:我本来的想法是用单调栈求出每个点的管辖区间,然后问题就变成了巨麻烦的线段覆盖问题,就爆炸写了
一晚上假算法。正解就是枚举一个端点,然后二分找右端点的区间,因为满足一个很神奇的单调性,然后st表维护
一下区间最值就好了。
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define pii pair<int, int> using namespace std; const int N = 2e5 + 7; const int M = 1e5 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 420047; int n, a[N], b[N], bin[20], Log[N], mx[20][N], mn[20][N]; void calRmq() { bin[0] = 1; Log[0] = -1; for(int i = 1; i < 20; i++) bin[i] = bin[i - 1] * 2; for(int i = 1; i < N; i++) Log[i] = Log[i / 2] + 1; for(int i = 1; i <= n; i++) mx[0][i] = a[i]; for(int i = 1; i <= n; i++) mn[0][i] = b[i]; for(int i = 1; i <= Log[n]; i++) { for(int j = 1; j <= n; j++) { if(j + bin[i] - 1 <= n) { mn[i][j] = min(mn[i - 1][j], mn[i - 1][j + bin[i - 1]]); mx[i][j] = max(mx[i - 1][j], mx[i - 1][j + bin[i - 1]]); } } } } int getVal(int x, int y, int op) { int t = Log[y - x + 1]; if(op) return max(mx[t][x], mx[t][y - bin[t] + 1]); return min(mn[t][x], mn[t][y - bin[t] + 1]); } int main(){ scanf("%d", &n); for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); } for(int i = 1; i <= n; i++) { scanf("%d", &b[i]); } calRmq(); LL ans = 0; for(int i = 1; i <= n; i++) { int l = i, r = n, mid, ret1 = -1, ret2 = -1; while(l <= r) { mid = l + r >> 1; int valMin = getVal(i, mid, 0); int valMx = getVal(i, mid, 1); if(valMin > valMx) l = mid + 1; else if(valMin < valMx) r = mid - 1; else ret1 = mid, r = mid - 1; } l = i, r = n; while(l <= r) { mid = l + r >> 1; int valMin = getVal(i, mid, 0); int valMx = getVal(i, mid, 1); if(valMin > valMx) l = mid + 1; else if(valMin < valMx) r = mid - 1; else ret2 = mid, l = mid + 1; } if(ret1 != -1) { ans += ret2 - ret1 + 1; } } printf("%lld\n", ans); return 0; } /* */
#include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define pii pair<int, int> using namespace std; const int N = 1e5; const int M = 1e6 + 7; const int inf = 0x3f3f3f3f; const LL INF = 0x3f3f3f3f3f3f3f3f; const int mod = 1e9 +7; int mm[N]; struct ST { int dp[N][20],ty; void build(int n,int b[],int _ty) { ty=_ty; for(int i=1;i<=n;i++) dp[i][0]=ty*b[i]; for(int j=1;j<=mm[n];j++) for(int i=1;i+(1<<j)-1<=n;i++) dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]); } int query(int x,int y) { int k=mm[y-x+1]; return ty*max(dp[x][k],dp[y-(1<<k)+1][k]); } }rmq; int a[N]; int main() { for(int i=-(mm[0]=-1);i<N;i++) mm[i]=mm[i-1]+((i&(i-1))==0); int n; scanf("%d", &n); for(int i = 1; i <= n; i++) { scanf("%d", &a[i]); } rmq.build(n, a, 1); while(1) { int x, y; scanf("%d%d", &x, &y); cout << rmq.query(x, y) << endl; } return 0; }
还有一个很神奇的st表
Codeforces Round #361 (Div. 2) D - Friends and Subsequences
标签:st表 seq const printf codeforce str class long mes
原文地址:https://www.cnblogs.com/CJLHY/p/9233358.html