To Be Or Not To Be
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
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
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
3
5 3
3 3 3
10 11 2
7 4 12
4 2 9
T 1
T 2
A 7
T 3
C
6 3
2 2 2
1 4
5 2
4 2
T 2
T 1
L
A 2
T 3
C
5 3
2 2 2
1 3
4 2
5 2
T 2
T 1
E 3
A 1
L
Sample Output
9:0
0:5
1:2
I will be back!!
斜堆模板題， 和左偏差不多
（允許我懶一發。。）
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
const int MAXN = 5000000;
namespace mheap {
const int MAXD = MAXN;
struct Node {
int w, l, r, fa;
} d[MAXD];
int tot = 0;
inline int newNode (int val = 0) {
memset(d+(++tot), 0, sizeof d[0]);
d[tot].w = val;
return tot;
}
inline void clear() {
tot = 0;
}
inline int merge(int a, int b) {
if(a == 0 || b == 0) return a+b;
if(d[a].w < d[b].w) std :: swap(a, b);
d[a].l = merge(d[a].l, b);
std :: swap(d[a].l, d[a].r);
return a;
}
int que[MAXD];
int build(int * l, int * r) {
int lef = 0, rig = -1;
while(l+(++rig) != r)
que[rig] = newNode(l[rig]);
--rig;
while(lef != rig) {
int a = que[lef++];
lef %= MAXD;
int b = que[lef++];
lef %= MAXD;
++rig; rig%=MAXD;
que[rig] = merge(a, b);
//printf("mg:%d %d\n", a, b);
}
return que[rig];
}
inline int pop(int u) {
return merge(d[u].l, d[u].r);
}
inline int & val(int u) {
return d[u].w;
}
int size(int u) {
if(!u) return 0;
return size(d[u].l) + size(d[u].r) + 1;
}
}
int sz[MAXN];
int buf[MAXN];
int piles[MAXN];
int player[2];
int main () {
int r;
while(~scanf("%d", &r)) {
mheap :: clear();
int cnt1 = 0, cnt2 = 0;
for(int t = 0; t<r; t++) {
player[0] = player[1] = 0;
int n, m;
scanf("%d%d", &n, &m);
for(int i = 1; i<=m; i++)
scanf("%d", &sz[i]);
for(int i = 1; i<=m; i++) {
for(int j = 1; j<=sz[i]; j++)
scanf("%d", &buf[j]);
piles[i] = mheap :: build(buf+1, buf+sz[i]+1);
}
char ord[5];
int arg;
for(int i = 0; i<n; i++) {
int & p1 = player[i&1], & p2 = player[!(i&1)];
scanf("%s", ord);
if(ord[0] == 'T') {
scanf("%d", &arg);
p1 = mheap :: merge(piles[arg], p1);
}
else if(ord[0] == 'C') {
//printf("p1:%d p2:%d\n", mheap :: val(p1), mheap :: val(p2));
if(mheap :: val(player[1]) > mheap :: val(player[0]))
player[1] = mheap :: merge(player[1], player[0]), player[0] = 0;
else if(mheap :: val(player[1]) < mheap :: val(player[0]))
player[0] = mheap :: merge(player[1], player[0]), player[1] = 0;
}
else if(ord[0] == 'L')
p1 = mheap :: pop(p1);
else if(ord[0] == 'A') {
scanf("%d", &arg);
mheap :: val(p1) += arg;
}
else if(ord[0] == 'E') {
scanf("%d", &arg);
p1 = mheap :: pop(p1);
p1 = mheap :: merge(p1, mheap :: newNode(arg));
}
}
int ans1 = mheap :: size(player[0]);
int ans2 = mheap :: size(player[1]);
printf("%d:%d\n", ans1, ans2);
if(ans1 > ans2) cnt1 ++;
else cnt2 ++;
}
if(cnt1 > cnt2) puts("Hahaha...I win!!");
else puts("I will be back!!");
}
}