題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=6638
題目大意：在二維平面空間有 n 個點，每個點有一個點權 wi，一個邊平行于坐標軸的矩形的權值為該矩形內所有點的點權和。問選取一個矩形點權和最大是多少。
題解：先將坐標離散化，以x坐標建線段樹，枚舉y坐標上界和下界，將y坐標位于上下界內的點加入線段樹，每次查詢線段樹最大子段和。
#include<bits/stdc++.h>
using namespace std
;
const int maxn
= 1e4 + 10;
typedef long long ll
;
struct ss
{int l
,r
;ll lv
,rv
,mv
,sum
;
}tree
[maxn
<< 2];
void pushup(int rt
){tree
[rt
].lv
= max(tree
[rt
<< 1].lv
,tree
[rt
<< 1].sum
+ tree
[rt
<< 1 | 1].lv
);tree
[rt
].rv
= max(tree
[rt
<< 1 | 1].rv
,tree
[rt
<< 1 | 1].sum
+ tree
[rt
<< 1].rv
);tree
[rt
].mv
= max(max(tree
[rt
<< 1].mv
,tree
[rt
<< 1 | 1].mv
),tree
[rt
<< 1].rv
+ tree
[rt
<< 1 | 1].lv
);tree
[rt
].sum
= tree
[rt
<< 1].sum
+ tree
[rt
<< 1 | 1].sum
;
}
void build(int rt
,int l
,int r
){tree
[rt
].l
= l
;tree
[rt
].r
= r
;if(tree
[rt
].l
== tree
[rt
].r
){tree
[rt
].lv
= tree
[rt
].rv
= tree
[rt
].mv
= tree
[rt
].sum
= 0;return ;}int mid
= l
+ r
>> 1;build(rt
<< 1,l
,mid
);build(rt
<< 1 | 1,mid
+ 1,r
);pushup(rt
);
}
void update(int rt
,int p
,ll k
){if(tree
[rt
].l
== tree
[rt
].r
){tree
[rt
].lv
+= k
;tree
[rt
].rv
+= k
;tree
[rt
].mv
+= k
;tree
[rt
].sum
+= k
;return ;}int mid
= tree
[rt
].l
+ tree
[rt
].r
>> 1;if(p
> mid
) update(rt
<< 1 | 1,p
,k
);else update(rt
<< 1,p
,k
);pushup(rt
);
}
ss
query(int rt
,int l
,int r
){if(tree
[rt
].l
>= l
&& tree
[rt
].r
<= r
) return tree
[rt
];int mid
= tree
[rt
].l
+ tree
[rt
].r
>> 1;if(l
<= mid
&& r
> mid
){ ss x1
= query(rt
<< 1,l
,mid
);ss x2
= query(rt
<< 1 | 1,mid
+ 1,r
);ss ans
;ans
.lv
= max(x1
.lv
,x1
.sum
+ x2
.lv
);ans
.rv
= max(x2
.rv
,x2
.sum
+ x1
.rv
);ans
.mv
= max(max(x1
.mv
,x2
.mv
),x1
.rv
+ x2
.lv
);ans
.sum
= x1
.sum
+ x2
.sum
;return ans
;}else{if(l
> mid
) return query(rt
<< 1 | 1,l
,r
);else return query(rt
<< 1,l
,r
);}
}
int t
,n
;
int x
[maxn
],y
[maxn
],z
[maxn
];
int tx
[maxn
],ty
[maxn
],px
,py
;
vector
<int> g
[maxn
];
int main() {scanf("%d",&t
);while(t
--) {px
= py
= 0;scanf("%d",&n
);for(int i
= 1; i
<= n
; i
++) {scanf("%d%d%d",&x
[i
],&y
[i
],&z
[i
]);tx
[i
] = x
[i
];ty
[i
] = y
[i
];g
[i
].clear();}sort(tx
+ 1,tx
+ n
+ 1,less
<int>());sort(ty
+ 1,ty
+ n
+ 1,less
<int>());px
= unique(tx
+ 1,tx
+ n
+ 1) - tx
- 1;py
= unique(ty
+ 1,ty
+ n
+ 1) - ty
- 1;for(int i
= 1; i
<= n
; i
++) {x
[i
] = lower_bound(tx
+ 1,tx
+ px
+ 1,x
[i
]) - tx
;y
[i
] = lower_bound(ty
+ 1,ty
+ py
+ 1,y
[i
]) - ty
;}for(int i
= 1; i
<= n
; i
++)g
[y
[i
]].push_back(i
);ll ans
= 0;for(int i
= 1; i
<= py
; i
++) {build(1,1,px
);for(int j
= i
; j
<= py
; j
++) {for(auto k
: g
[j
])update(1,x
[k
],z
[k
]);ss res
= query(1,1,px
);ans
= max(ans
,res
.mv
);}}printf("%lld\n",ans
);}return 0;
}
總結
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