郁闷的出纳员   splayTree

测试数据地址：http://115.com/file/aqv0qbbp



郁闷的出纳员

题目: http://www.zybbs.org/JudgeOnline/problem.php?id=1503

题目大意：见题目吧，
题意很明确，
就是补充一点，相同的工资是不能合并的，比如说

有11，
11，
11，那么第一是11，

第三也是11

思路：也没什么思路，
就是为了测试splayTree模板，可能有点儿技巧的就是两个地方，一是在全体增加和全体减少的时候，用一个东西del存下来，将数字x放入树时，放入x-del

这样对于每次的增加操作就不用遍历树了，只需在减少的时候进行删除操作。原理也很简单。

第二是在删除的时候，如果一个节点rt不满足了，直接将其及左树，将右树接到rt父节点处。



提交情况：CE。。。-——》wa——》
TLE—》WA
——》AC



AC
code：







#include
<cstdio>

#include
<cstring>



#define
maxn 1000000



struct
splayTreeNode{


int key, size, id;


splayTreeNode* son[2],* father;


void init(int k, int sz, int chi, splayTreeNode* l, splayTreeNode*
r, splayTreeNode* fa){


key = k, size = sz, id = chi, son[0] = l, son[1] = r, father =
fa;


}

};



struct
splayTree{


splayTreeNode* nul, *link;




#define root nul->son[0]


int ad;


splayTree(){


link = new splayTreeNode[maxn];


ad = 0;


nul = &(link[ad ++]);


nul->init(0, 0, 0, NULL, NULL, NULL);


}




void rotate(splayTreeNode* &rt, int son1, int
son2){


splayTreeNode* temp = rt->son[son1];


rt->son[son1] =
temp->son[son2];


rt->size -= temp->size;


if(temp->son[son2] != NULL){


temp->size -=
temp->son[son2]->size;


rt->size +=
temp->son[son2]->size;


temp->son[son2]->father =
rt;


temp->son[son2]->id =
son1;


}


temp->son[son2] = rt;


temp->father = rt->father;


temp->id = rt->id;


temp->size += rt->size;


rt->father = temp;


rt->id = son2;


rt = temp;


}




void splay(splayTreeNode* x, splayTreeNode* rt){


if(x->father == rt) return;


splayTreeNode* y = x->father,* z =
x->father->father;


if(z == rt){


if(y->son[0] == x)
rotate(z->son[y->id], 0,
1);


else rotate(z->son[y->id], 1,
0);


}


else{


if(y->id == x->id){


rotate(z->father->son[z->id],
x->id, (x->id)^1);


rotate(y->father->son[y->id],
x->id, (x->id)^1);


}


else{


rotate(y->father->son[y->id],
x->id, (x->id)^1);


rotate(z->father->son[z->id],
y->id, (y->id)^1);


}


}


splay(x, rt);


}




void insert(int x, int chi, splayTreeNode* &rt,
splayTreeNode* rf){


if(rt == NULL){


rt = &link[ad ++];


if(rf == nul) nul->son[0] = root;


rt->init(x, 1, chi, NULL, NULL, rf);


splay(rt, this->nul);


return;


}


rt->size ++;


if(x < rt->key)
insert(x, 0, rt->son[0], rt);


else insert(x, 1, rt->son[1], rt);


}


void rise(int x, splayTreeNode* rt){


if(rt == NULL) return;


rt->key += x;


rise(x, rt->son[0]);


rise(x, rt->son[1]);


}


void reduce(int x, int min, splayTreeNode* rt){


if(rt == NULL) return;


if(rt->key + x < min){


rt->father->son[rt->id]
= rt->son[1];


if(rt->son[1] != NULL){


rt->son[1]->id =
rt->id;


rt->son[1]->father =
rt->father;


reduce(x, min, rt->son[1]);


}


}


else{


reduce(x, min, rt->son[0]);


rt->size = 1;


if(rt->son[1] != NULL) rt->size +=
rt->son[1]->size;


if(rt->son[0] != NULL) rt->size +=
rt->son[0]->size;


}


}


int find(int k, splayTreeNode* rt){


if(rt == NULL || rt->size < k) return
-1;


if(rt->son[1] != NULL){


if(rt->son[1]->size == k - 1) return
rt->key;


if(rt->son[1]->size >=
k) return find(k, rt->son[1]);


return find(k - (rt->son[1]->size +
1), rt->son[0]);


}


else{


if(k == 1) return rt->key;


else return find(k - 1, rt->son[0]);


}


}

};



int
main(){




char ord[3];


int k, n, min, tot, del;


while(~scanf("%d %d", &n,
&min)){


splayTree spl;


del = tot = 0;


for(int i = 0; i < n; ++ i){


scanf("%1s %d", ord, &k);


switch(ord[0]){


case 'I': if(k
>= min){


tot ++;


spl.insert(k - del, 0, spl.root, spl.nul);


}


break;


case 'A': del
+= k;


break;


case 'S':
del -= k;


spl.reduce(del, min, spl.root);


break;


case 'F':
int ans = spl.find(k, spl.root);


printf("%d\n", ans == -1 ? ans : (ans + del));


break;


}


}


printf("%d\n", tot - spl.root->size);


}


return 0;

}