2018CCPC网络赛1010 HDU6447 YJJ's Salesman（线段树 + DP + 离散化）

YJJ's Salesman Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)   Problem Description YJJ is a salesman who has traveled through western country. YJJ is always on journey. Either is he at the destination, or on the way to destination. One day, he is going to travel from city A to southeastern city B. Let us assume that A is (0,0) on the rectangle map and B (109,109) . YJJ is so busy so he never turn back or go twice the same way, he will only move to east, south or southeast, which means, if YJJ is at (x,y) now (0≤x≤109,0≤y≤109) , he will only forward to (x+1,y) , (x,y+1) or (x+1,y+1) . On the rectangle map from (0,0) to (109,109) , there are several villages scattering on the map. Villagers will do business deals with salesmen from northwestern, but not northern or western. In mathematical language, this means when there is a village k on (xk,yk) (1≤xk≤109,1≤yk≤109) , only the one who was from (xk−1,yk−1) to (xk,yk) will be able to earn vk dollars.(YJJ may get different number of dollars from different village.) YJJ has no time to plan the path, can you help him to find maximum of dollars YJJ can get.     Input The first line of the input contains an integer T (1≤T≤10) ,which is the number of test cases. In each case, the first line of the input contains an integer N (1≤N≤105) .The following N lines, the k -th line contains 3 integers, xk,yk,vk (0≤vk≤103) , which indicate that there is a village on (xk,yk) and he can get vk dollars in that village. The positions of each village is distinct.     Output The maximum of dollars YJJ can get.     Sample Input   1 3 1 1 1 1 2 2 3 3 1     Sample Output   3

 

题目大意是说在一个1e9*1e9的地图上有1e5个村庄，每个村庄有一定的利润。每次只能从(x,y)走到(x+1,y)或(x,y+1)或(x+1,y+1)，并且只有是用最后一种走法走到村庄的才能获得该村庄的利益。

我的做法是先把村庄按照纵坐标从小到大，横坐标从大到小排序一下（横坐标从小到大是类似01背包），再把横坐标离散化一下这样遍历村庄的时候就能直接在一个一维的数组里找最大利益并更新当前横坐标的利益，用线段树或树状数组维护最大利益即可。

#include <bits/stdc++.h>
using namespace std;
const int maxn = 100010;
typedef long long ll;
int maxx[4 * maxn];
int n;
int x[maxn];
struct fun
{
    int x,y,v;
}z[maxn];
int cntx;
void update(int l, int r, int rt, int p, int v)
{
    if (p <= l && r <= p)
    {
        maxx[rt] = max(maxx[rt], v);
        return;
    }
    int mid = (l + r) >> 1;
    if (p <= mid)
        update(l, mid, rt << 1, p, v);
    else
        update(mid + 1, r, (rt << 1) + 1, p, v);
    maxx[rt] = max(maxx[rt << 1], maxx[(rt << 1) + 1]);
}
int query(int l, int r, int rt, int l1, int r1)
{
    if (l > r || l1 > r1)
        return 0;
    if (l1 <= l && r <= r1)
        return maxx[rt];
    int mid = (l + r) >> 1;
    int ans1,ans2;
    ans1 = ans2 = 0;
    if (l1 <= mid)
        ans1 = query(l, mid, rt << 1, l1, r1);
    if (r1 > mid)
        ans2 = query(mid + 1, r, (rt << 1) + 1, l1, r1);
    return max(ans1, ans2);
}
bool cmp(fun a, fun b)
{
    if (a.y != b.y)
        return a.y < b.y;
    return a.x > b.x;
}
int main()
{
//    freopen("in.txt", "r", stdin);
    int t;
    scanf("%d", &t);
    while (t--)
    {
        memset(maxx, 0, sizeof(maxx));
        cntx = 1;
        scanf("%d", &n);
        for (int i = 0; i < n; i++)
        {
            scanf("%d%d%d", &z[i].x, &z[i].y, &z[i].v);
            x[i] = z[i].x;
        }
        sort(x, x + n);
        for (int i = 0; i < n; i++)
            z[i].x = lower_bound(x, x + n, z[i].x) - x + 1;
        sort(z, z + n, cmp);
        int ans = 0;
        int tmp;
        for (int i = 0; i < n; i++)
        {
            tmp = query(1, n, 1, 1, z[i].x - 1) + z[i].v;
            ans = max(ans, tmp);
            update(1, n, 1, z[i].x, tmp);
        }
        printf("%d\n", ans);
    }
    return 0;
}