FZU2013 A short problem —— 线段树/树状数组 + 前缀和

题目链接：https://vjudge.net/problem/FZU-2013

 Problem 2013 A short problem

Accept: 356    Submit: 1083
Time Limit: 1000 mSec    Memory Limit : 32768 KB

 Problem Description

The description of this problem is very short. Now give you a string(length N), and ask you the max sum of the substring which the length can't small than M.

 Input

The first line is one integer T(T≤20) indicates the number of the test cases. Then for every case, the first line is two integer N(1≤N≤1000000) and M(1≤M≤N).

Then one line contains N integer indicate the number. All the number is between -10000 and 10000.

 Output

Output one line with an integer.

 Sample Input

2
5 1
1 -2 -2 -2 1
5 2
1 -2 -2 -2 1

 Sample Output

1
-1

 Source

FOJ有奖月赛-2011年03月

题意：

给出一个序列，求长度不小于m且和最大的子序列。

题解：

1.求出前缀和。

2.将0插入到线段树第0位，然后从第m为开始枚举：去线段树范围为0~i-m的地方找最小值，然后以i为结尾的子序列最大值即为：sum[i] - query(1,0,n,0,i-m)。答案即为max（sum[i] - query(1,0,n,0,i-m)）。

3.类似的题目：CSU - 1551 Longest Increasing Subsequence Again

线段树：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 1e6;

 int minv[MAXN*];
 void add(int u, int l, int r, int x, int val)
 {
     if(l==r)
     {
         minv[u] = val;
         return;
     }
     int mid = (l+r)/;
     if(x<=mid) add(u*, l, mid, x, val);
     else add(u*+, mid+,r, x, val);
     minv[u] = min(minv[u*], minv[u*+]);
 }

 int query(int u, int l, int r, int x, int y)
 {
     if(l<=x&&y<=r)
         return minv[u];

     int mid = (l+r)/;
     int ret = INF;
     if(x<=mid) ret = min(ret, query(u*, l, mid, x, mid));
     if(y>=mid+) ret = min(ret, query(u*+, mid+, r, mid+, y));
     return ret;
 }

 int sum[MAXN];
 int main()
 {
     int T, n, m;
     scanf("%d", &T);
     while(T--)
     {
         scanf("%d%d", &n,&m);
         sum[] = ;
         for(int i = ; i<=n; i++)
         {
             int val;
             scanf("%d", &val);
             sum[i] = sum[i-]+val;
         }

         for(int i = ; i<=n*; i++)
             minv[i] = INF;

         int ans = -INF;
         add(,,n,,);
         for(int i = m; i<=n; i++)
         {
             ans = max(ans, sum[i]-query(,,n,,i-m));
             add(,,n,i-m+,sum[i-m+]);
         }
         printf("%d\n", ans);
     }
 }

树状数组：

 #include <iostream>
 #include <cstdio>
 #include <cstring>
 #include <algorithm>
 #include <vector>
 #include <cmath>
 #include <queue>
 #include <stack>
 #include <map>
 #include <string>
 #include <set>
 using namespace std;
 typedef long long LL;
 const int INF = 2e9;
 const LL LNF = 9e18;
 const int MOD = 1e9+;
 const int MAXN = 1e6;

 int T, n, m;
 int lowbit(int x)
 {
     return x&(-x);
 }

 int c[MAXN];
 void add(int x, int d)
 {
     while(x<=n)
     {
         c[x] = min(c[x], d);
         x += lowbit(x);
     }
 }

 int query(int x)
 {
     int s = INF;
     while(x>)
     {
         s = min(c[x], s);
         x -= lowbit(x);
     }
     return s;
 }

 int sum[MAXN];
 int main()
 {
     scanf("%d", &T);
     while(T--)
     {
         memset(c, , sizeof(c));
         scanf("%d%d", &n,&m);
         sum[] = ;
         for(int i = ; i<=n; i++)
         {
             int val;
             scanf("%d", &val);
             sum[i] = sum[i-]+val;
         }
         for(int i = ; i<=n+; i++)
             c[i] = INF;

         int ans = -INF;
         add(, );
         for(int i = m; i<=n; i++)
         {
             ans = max(ans, sum[i]-query(i-m+));
             add(i-m+, sum[i-m+]);
         }
         printf("%d\n", ans);
     }
 }