HDU3031及左偏树模板

////  main.cpp//  leftist-tree////  Created by 蘇與軒 on 15/1/28.//  Copyright (c) 2015年 蘇與軒. All rights reserved.//#include <iostream>#include <cstdio>#include <cstring>#include <queue>#include <vector>#include <cmath>#include <cstdlib>#include <string>#include <map>#include <set>#include <algorithm>#include <functional>#define rep(i,a,b) for (int i=a;i<(b+1);i++)#define Rep(i,a,b) for (int i=a;i>=b;i--)#define mid ((l+r)>>1)#define lson (k<<1)#define rson (k<<1|1)using namespace std;#define M 1000010struct P{    int l,r,dis,v;}tree[M];int f[110];int Merge(int x,int y) {    if (!x) return y;    if (!y) return x;    if (tree[x].v<tree[y].v)    swap(x,y);//big treap    tree[x].r=Merge(tree[x].r,y);    if (tree[tree[x].l].dis<tree[tree[x].r].dis)    swap(tree[x].l,tree[x].r);    if (tree[x].r)  tree[x].dis=tree[tree[x].r].dis+1;    else tree[x].dis=0;    return x;}int Pop(int x) {    int l=tree[x].l;    int r=tree[x].r;    tree[x].l=tree[x].r=tree[x].dis=0;    return Merge(l,r);}int T,n,m,p[110],pp,x,root[2],sum[110],tot,sheep,wolf;char s[5];int main(int argc, const char * argv[]) {    scanf("%d",&T);    sheep=0;wolf=0;    while (T--) {        scanf("%d%d",&n,&m);        memset(f,0,sizeof f);        rep(i,1,m)  scanf("%d",&p[i]);        int tot=1;        rep(i,1,m)  rep(j,1,p[i]){            scanf("%d",&pp);            tree[tot].l=tree[tot].r=tree[tot].dis=0;            tree[tot].v=pp;            f[i]=Merge(f[i],tot);            tot++;        }        memset(sum,0,sizeof sum);        root[0]=root[1]=0;        rep(i,0,n-1) {            scanf("%s",s);            switch (s[0]) {                case 'T':                    scanf("%d",&x);                    root[i&1]=Merge(f[x],root[i&1]);                    sum[i&1]+=p[x];                    f[x]=0;                    p[x]=0;                    break;                case 'C':                    if (tree[root[0]].v>tree[root[1]].v){                        root[0]=Merge(root[0],root[1]);                        sum[0]+=sum[1];                        root[1]=0;                        sum[1]=0;                    }else{                        root[1]=Merge(root[0],root[1]);                        sum[1]+=sum[0];                        root[0]=0;                        sum[0]=0;                    }                    break;                case 'L':                    root[i&1]=Pop(root[i&1]);                    sum[i&1]--;                    break;                case 'A':                    scanf("%d",&x);                    tree[root[i&1]].v+=x;                    break;                case 'E':                    scanf("%d",&x);                    int cur=Pop(root[i&1]);                    tree[cur].v=x;                    root[i&1]=Merge(cur,root[i&1]);                    break;            }        }        printf("%d:%d\n",sum[0],sum[1]);    }    if (sum[0]>sum[1]) {        wolf++;    }else sheep++;    if (sheep>wolf) printf("I will be back!!\n");    else printf("Hahaha...I win!!\n");    return 0;}

关于左偏树的教程很多大牛的博客和网上都有，就不累述了，直接上代码！

题意：

Problem Description

That's a question. Now Happy (Xi Yangyang) has been caught by Wolffy (Hui Tailang). As Wolffy is busy preparing the big meal, a good idea comes to Happy. He proposes a game that only Wolffy had won, he can eat Happy. Wolffy always believes he is the cleverest one, so they reach a consensus. And they both agree with Wolnie (Hong Tailang) when the referee. A theater will be beat to die by Wolnie's pan.


The game is defined as follow.

There are multiple test cases.

In each case there are R (R < 10) rounds of the game, R is an odd number to guarantee that there must be a winner in the end.

In each round: There is a pile of n (10 <= n <= 200) Special-cards and m (1 <= m <= 100) piles of Point-card on the table. The Point-card piles are ordered from 1 to m. Wolffy and Happy take turns to get one card from the top of Special-cards pile. Wolffy always takes first in the game. When all the Special-cards have been taken, the round is over and the one with more cards in the hands gains one point. If there is a tie, Wolffy gains one point.(Wolffty and Happy both have 0 point before the game).

There are 5 kinds of Special-cards besides the Point-card in the game.

0) Point-card: a card with a point X (1 <= X <= 2000000).

1) Challenge-card: no matter who takes this card, they both take one card with the maximum point from their own hands. After a comparison, if Happy's card has a larger point, He takes all the Wolffy's in-hands cards, vice versa; If there is a tie no more operation.

2) Loss-card: the one who takes this card, He must throw a card with the maximum point.

3) Add-card: a card with P point, the one who gets this card will make the card with maximum point P point larger, i.e. if a Point-card with X point is the maximum, its point will change to X + P. An Add-card can only work on one Point-card.

4) Exchange-card: a card with Q point. The one who gets this card must change one maximum-point card's point to Q.

5) Take-card: a card with a integer K, indicates one can get the all the cards of Kth Point-card pile. In one round no two Take-card have the same K.

You can assume that when one gets the Loss-card, Add-card, Exchange-card, He has at least one card in the hands, when one gets a Challenge-card, they both have at least one card in the hands.

 

Input

Input

For each test case, the first line of input is an integer R, indicates the number of rounds:

Line 2: two integers n indicates the number of Special-cards, m indicates the number of Point-card piles.

Line 3: a line of m integers. The ith number Pi (1 <= Pi <= 10000)indicates the number cards of ith Point-card pile.

For the next m lines, ith line contains Pi numbers indicate every Point-card's point of ith Point-card pile.

The next n lines, in each line, there are five kinds of input, indicate Special-cards by the order of "from top to bottom".

1) T K: indicates one gets a Take-card, and He can get Kth Point-card pile(1 <= K <= m).

2) C: indicates one gets a Challenge card.

3) L: indicates one gets a Loss card.

4) A P: indicates one gets an Add card with P point (1 <= P <= 30).

5) E Q: indicates one gets an Exchange card with Q point (1 <= Q <= 2000000).

 

Output

For each round you should print A:B in a line. A indicate the number of left cards of Wolffy, B indicates the number of left cards of Happy. At the end of game, if Wolffy gains more points, print "Hahaha...I win!!", else print "I will be back!!".

 

Sample Input

35 33 3 310 11 27 4 124 2 9T 1T 2A 7T 3C6 32 2 21 45 24 2T 2T 1LA 2T 3C5 32 2 21 34 25 2T 2T 1E 3A 1L

 

Sample Output

9:00:51:2I will be back!!

 

Source

2009 Multi-University Training Contest 13 - Host by HIT 

 

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