标签：acm   hdu   

Magician

Fantasy magicians usually gain their ability through one of three usual methods: possessing it as an innate talent, gaining it through study and practice, or receiving it from another being, often a god, spirit, or demon of some sort. Some wizards are depicted as having a special gift which sets them apart from the vast majority of characters in fantasy worlds who are unable to learn magic.

Magicians, sorcerers, wizards, magi, and practitioners of magic by other titles have appeared in myths, folktales, and literature throughout recorded history, with fantasy works drawing from this background.

In medieval chivalric romance, the wizard often appears as a wise old man and acts as a mentor, with Merlin from the King Arthur stories representing a prime example. Other magicians can appear as villains, hostile to the hero.



Mr. Zstu is a magician, he has many elves like dobby, each of which has a magic power (maybe negative). One day, Mr. Zstu want to test his ability of doing some magic. He made the elves stand in a straight line, from position 1 to position n, and he used two kinds of magic, Change magic and Query Magic, the first is to change an elf’s power, the second is get the maximum sum of beautiful subsequence of a given interval. A beautiful subsequence is a subsequence that all the adjacent pairs of elves in the sequence have a different parity of position. Can you do the same thing as Mr. Zstu ?

1
1 1
1
0 1 1

1

2015 Multi-University Training Contest 3

简单的线段树区间叠加，维护区间内以奇偶开头/结尾的最大值，注意初始化时全部初始化为-INF， 在更新的时候就会选择最大的进行更新，自动的决定选取或者不选取。

#include<cstdio>
#include<cmath>
#include<stdlib.h>
#include<map>
#include<set>
#include<time.h>
#include<vector>
#include<queue>
#include<string>
#include<string.h>
#include<iostream>
#include<algorithm>
using namespace std;
#define eps 1e-8
#define INF 0x3f3f3f3f
#define max(a,b) ((a)>(b)?(a):(b))
#define min(a,b) ((a)<(b)?(a):(b))
#define lson L, mid, rt<<1
#define rson mid+1, R, rt<<1|1
#define maxn 100000 + 10

long long a[maxn];
int n, m;

struct Node
{
    long long M[2][2];
}T[maxn<<2];

void pushup(int rt)
{
    int l = rt<<1, r = rt<<1|1;
    for(int i=0; i<2; i++)
    for(int j=0; j<2; j++)
    {
        T[rt].M[i][j] = max(max(T[l].M[i][j], T[r].M[i][j]), max(T[l].M[i][0]+T[r].M[1][j], T[l].M[i][1]+T[r].M[0][j]));
    }
}

void build(int L, int R, int rt)
{
    if(L == R)
    {
        scanf("%I64d", &a[L]);
        if(L & 1)
        {
            T[rt].M[1][1] = a[L];
            T[rt].M[0][0] = -INF;
        }
        else
        {
            T[rt].M[0][0] = a[L];
            T[rt].M[1][1] = -INF;
        }
            T[rt].M[1][0] = -INF;
            T[rt].M[0][1] = -INF;
        return ;
    }
    int mid = (L + R) >> 1;
    build(lson);
    build(rson);
    pushup(rt);
}

void update(int t, int v, int L, int R, int rt)
{
    if(L == R)
    {
        if(L & 1)
        {
            T[rt].M[1][1] = v;
        }
        else
        {
            T[rt].M[0][0] = v;
        }
        return ;
    }
    int mid = (L + R) >> 1;
    if(t <= mid) update(t, v, lson);
    else update(t, v, rson);
    pushup(rt);

}

Node query(int l, int r, int L, int R, int rt)
{
    Node tmp;
    if(l == L && r == R)
    {
        tmp = T[rt];
        return tmp;
    }
    int mid = (L + R) >> 1;
    if(r <= mid) return query(l, r, lson);
    else if(l > mid) return query(l, r, rson);
    else
    {
        Node ltmp = query(l, mid, lson);
        Node rtmp = query(mid+1, r, rson);
        for(int i=0; i<2; i++)
        for(int j=0; j<2; j++)
        {
            tmp.M[i][j] = max(max(ltmp.M[i][j], rtmp.M[i][j]), max(ltmp.M[i][0]+rtmp.M[1][j], ltmp.M[i][1]+rtmp.M[0][j]));
        }
        return tmp;
    }
}

int main()
{
    int kase;
    scanf("%d", &kase);
    while(kase--)
    {
        scanf("%d%d", &n, &m);
        build(1, n, 1);

        for(int i=0; i<m; i++)
        {
            int t, a, b;
            scanf("%d%d%d", &t, &a, &b);
            if(t == 0)
            {
                long long ans = -INF;
                Node tmp = query(a, b, 1, n, 1);
                for(int i=0; i<2; i++)
                for(int j=0; j<2; j++)
                {
                    ans = max(ans, tmp.M[i][j]);
                }
                printf("%I64d\n", ans);
            }
            else
            {
                update(a, b, 1, n, 1);
            }
        }
    }
}

hdu 多校

标签：acm   hdu   

原文地址：http://blog.csdn.net/dojintian/article/details/47125859