poj2479(子串和变形,DP)

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题目链接：poj2479

话说，一开始是用线段树A的。。。。。

/*题意：求数列中两个不重叠的子序列的最大和。思路：先从左向右求最大子串和(d1表示)，然后再从右向左求最大子串和(d2表示)然后遍历d1，记录1~i之间的最大值left，比较left和d2[i+1]的最大值*/#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define lson rt<<1#define rson rt<<1|1const int inf = 0x3f3f3f3f;const int N = 50005;int d1[N],d2[N],a[N];int main(){    int T,i,n;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(i = 1; i <= n; i ++)            scanf("%d",&a[i]);        memset(d1, 0, sizeof(d1));        memset(d2, 0, sizeof(d2));        for(i = 1; i <= n; i ++)//从左向右求最大子串和            d1[i] = max(a[i], a[i]+d1[i-1]);        for(i = n; i >= 1; i --)//从右向左求最大子串和            d2[i] = max(a[i], a[i]+d2[i+1]);        int ans = -inf, left = -inf;        for(i = 1; i < n; i ++){            left = max(left, d1[i]);            ans = max(ans, left+d2[i+1]);        }        printf("%d\n",ans);    }    return 0;}

线段树代码：

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;#define lson rt<<1#define rson rt<<1|1const int inf = 0x3f3f3f3f;const int N = 50005;int d1[N],d2[N],a[N];struct node{    int l,r;    int num;}s[N<<2];void build(int l, int r, int rt){    s[rt].l = l;  s[rt].r = r;    s[rt].num = -inf;    if(l == r){        s[rt].num = d2[l];        return;    }    int mid = (l + r) >> 1;    build(l, mid, lson);    build(mid+1, r, rson);    s[rt].num = max(s[lson].num, s[rson].num);}int query(int l, int r, int rt){    if(l <= s[rt].l && s[rt].r <= r)        return s[rt].num;    int mid = (s[rt].l + s[rt].r) >> 1;    int ans = -inf;    if(l <= mid)        ans = max(ans, query(l, r, lson));    if(mid < r)        ans = max(ans, query(l, r, rson));    return ans;}int main(){    int T,i,n;    scanf("%d",&T);    while(T--)    {        scanf("%d",&n);        for(i = 1; i <= n; i ++)            scanf("%d",&a[i]);        memset(d1, 0, sizeof(d1));        memset(d2, 0, sizeof(d2));        for(i = 1; i <= n; i ++)            d1[i] = max(a[i], a[i]+d1[i-1]);        for(i = n; i >= 1; i --)            d2[i] = max(a[i], a[i]+d2[i+1]);        build(1, n, 1);        int ans = -inf;        for(i = 1; i < n; i ++){            int num = query(i+1, n, 1);//在d2中查询i+1~n间的最大值            ans = max(ans, num+d1[i]);        }        printf("%d\n",ans);    }    return 0;}

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