HDU6447 YJJ's Salesman

树状数组 + 离散化

把坐标按y从小到大排序，y相同的按x从大到小排序，然后把x离散化以后用树状数组维护前缀最大值。（并不是严格的前缀最大值，对于之前树状数组中的最值，如果修改的值小于原值，仍然把原来的值当成最大值），就相当于选择了之前的点做转移。

这样保证了每次更新时，当前点一定是由左下方的矩阵转移而来。x要从小到大排序是因为保证每次更新对于y相同的点不会被其左边的点影响。

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define full(a, b) memset(a, b, sizeof a)
#define FAST_IO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int ret = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) ret = (ret << 3) + (ret << 1) + (ch ^ 48), ch = getchar();
    return w ? -ret : ret;
}
inline int gcd(int a, int b){ return b ? gcd(b, a % b) : a; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template <typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template <typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template <typename A, typename B, typename C>
inline A fpow(A x, B p, C lyd){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % lyd)if(p & 1)ans = 1LL * x * ans % lyd;
    return ans;
}
const int N = 200005;
int _, n, tot, a[N], tree[N], ans;
struct Pos{
    int x, y, w;
    bool operator < (const Pos &rhs) const {
        if(y != rhs.y) return y < rhs.y;
        return x > rhs.x;
    }
}pos[N];

inline void add(int k, int val){
    for(; k <= n; k += lowbit(k)) tree[k] = max(val, tree[k]);
}

inline int query(int k){
    int ret = 0;
    for(; k; k -= lowbit(k)) ret = max(ret, tree[k]);
    return ret;
}

int main(){

    for(_ = read(); _; _ --){
        n = read(), ans = 0;
        full(a, 0), full(tree, 0);
        for(int i = 1; i <= n; i ++){
            pos[i].x = read(), pos[i].y = read(), pos[i].w = read();
            a[++tot] = pos[i].x;
        }
        sort(pos + 1, pos + n + 1);
        sort(a + 1, a + tot + 1);
        tot = unique(a + 1, a + tot + 1) - a - 1;
        for(int i = 1; i <= n; i ++){
            int p = lower_bound(a + 1, a + tot + 1, pos[i].x) - a;
            int val = query(p - 1) + pos[i].w;
            ans = max(ans, val);
            add(p, val);
        }
        printf("%d
", ans);
    }
    return 0;
}