hdu5289 2015多校联合第一场1002 Assignment

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题意：给出一个数列，问其中存在多少连续子区间，其中子区间的(最大值-最小值)<k

思路：设dp[i]为从区间1到i满足题意条件的解，最终解即为dp[n]；

此外假设对于arr[i] 往左遍历一直到arr[r] 此时从区间r到区间i满足(最大值-最小值)<k，再往左一位即越界或者不满足条件，此时有 dp[i]= dp[i-1]+ i - r + 1;

因为数据量大往左遍历时可能会超时，所以用rmq打表查找r时用二分就过了

代码：

#include <algorithm>#include <iostream>#include <sstream>#include <cstdlib>#include <cstring>#include <iomanip>#include <cstdio>#include <string>#include <bitset>#include <vector>#include <queue>#include <stack>#include <cmath>#include <list>#include <map>#include <set>#define sss(a,b,c) scanf("%d%d%d",&a,&b,&c)#define mem1(a) memset(a,-1,sizeof(a))#define mem(a) memset(a,0,sizeof(a))#define ss(a,b) scanf("%d%d",&a,&b)#define s(a) scanf("%d",&a)#define INF 0x3f3f3f3f#define w(a) while(a)#define PI acos(-1.0)#define LL long long#define eps 10E-9#define N 100010using namespace std;void mys(int& res) {    int flag=0;    char ch;    while(!(((ch=getchar())>='0'&&ch<='9')||ch=='-'))        if(ch==EOF)  res=INF;    if(ch=='-')  flag=1;    else if(ch>='0'&&ch<='9')  res=ch-'0';    while((ch=getchar())>='0'&&ch<='9')  res=res*10+ch-'0';    res=flag?-res:res;}void myp(int a) {    if(a>9)       myp(a/10);    putchar(a%10+'0');}/********************the end of template********************/int arr[N];int sm[N][30], bg[N][30];LL dp[N];void rmq_init(int n) {    for(int i = 0; i < n + 1; i++)   bg[i][0] = sm[i][0] = arr[i];    for(int j = 1; (1 << j) <= n + 1; j++) {        for(int i = 0; i + (1 << j) - 1 < n + 1; i++) {            bg[i][j] = max(bg[i][j - 1], bg[i + (1 << (j - 1))][j - 1]);            sm[i][j] = min(sm[i][j - 1], sm[i + (1 << (j - 1))][j - 1]);        }    }}int rmq_min(int left, int right){    int kk = 0;    w((1 << (kk+1)) <= right - left +1) kk++;    return min(sm[left][kk], sm[right - (1<<kk) +1][kk]);}int rmq_max(int left, int right){    int kk = 0;    w((1 << (kk+1)) <= right - left +1) kk++;    return max(bg[left][kk], bg[right - (1<<kk) +1][kk]);}int getR(int L, int k, int R) {    int l = L, r = R, m, ans;    while(l <= r) {        m = (l + r) >> 1;        int view = rmq_max(m, R) - rmq_min(m, R);        if(view < k) {            ans = m;            r = m - 1;        }         else l = m + 1;    }    return ans;}int main(){    int t;    s(t);    w(t--){        int n, k;        mem(dp);        ss(n, k);        for(int i=1; i<=n; i++){            s(arr[i]);        }        rmq_init(n);        dp[1] = 1;        for(int i=2; i<=n; i++){            int R = getR(1, k, i);            dp[i] = dp[i-1] + i - R + 1;        }        cout<<dp[n]<<endl;    }    return 0;}

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