題目 ?：Gameia：鏈接

Time Limit: 2000/1000 MS (Java/Others)????Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 160????Accepted Submission(s): 53

Problem Description Alice and Bob are playing a game called 'Gameia ? Gameia !'. The game goes like this :
0. There is a tree with all node unpainted initial.
1. Because Bob is the VIP player, so Bob has K chances to make a small change on the tree any time during the game if he wants, whether before or after Alice's action. These chances can be used together or separate, changes will happen in a flash. each change is defined as cut an edge on the tree.
2. Then the game starts, Alice and Bob take turns to paint an unpainted node, Alice go first, and then Bob.
3. In Alice's move, she can paint an unpainted node into white color.
4. In Bob's move, he can paint an unpainted node into black color, and what's more, all the other nodes which connects with the node directly will be painted or repainted into black color too, even if they are white color before.
5. When anybody can't make a move, the game stop, with all nodes painted of course. If they can find a node with white color, Alice win the game, otherwise Bob.
Given the tree initial, who will win the game if both players play optimally?
Input The first line of the input gives the number of test cases T; T test cases follow.
Each case begins with one line with two integers N and K : the size of the tree and the max small changes that Bob can make.
The next line gives the information of the tree, nodes are marked from 1 to N, node 1 is the root, so the line contains N-1 numbers, the i-th of them give the farther node of the node i+1.

Limits
T≤100
1≤N≤500
0≤K≤500
1≤Pi≤i
Output For each test case output one line denotes the answer.
If Alice can win, output "Alice" , otherwise "Bob".
Sample Input 2 2 1 1 3 1 1 2
Sample Output Bob Alice 題解：- 如果Bob能把這棵樹分成若干兩個(gè)一組的點(diǎn)對，那么Bob取得勝利，否則Alice獲勝。

?- 如果原樹不存在兩兩匹配的方案，Alice從樹葉開始，每次都染樹葉父節(jié)點(diǎn)，Bob被迫只能不斷的染葉子，Bob退化成一般玩家，因?yàn)锽ob做不做小動(dòng)作都不會(huì)逆轉(zhuǎn)局勢，總會(huì)出現(xiàn)一個(gè)時(shí)間點(diǎn)Bob沒辦法跟上Alice的節(jié)奏而讓Alice染到一個(gè)周圍都已被染色的孤立點(diǎn)（因?yàn)樵瓨洳淮嬖趦蓛善ヅ涞姆桨?#xff09;?

- 如果原樹存在兩兩匹配的方案，而且Bob的小動(dòng)作次數(shù)也足以把原樹分成兩兩的點(diǎn)對，那么Bob顯然獲勝。 - 如果原樹存在兩兩匹配的方案，而Bob的小動(dòng)作不足以把樹分成兩兩的點(diǎn)對，Alice一定獲勝，因?yàn)槊看稳灸硞€(gè)葉子節(jié)點(diǎn)（該節(jié)點(diǎn)為其父節(jié)點(diǎn)的唯一子節(jié)點(diǎn)），Alice總能迫使Bob不斷的做小動(dòng)作以保證剩下的樹不會(huì)出現(xiàn)奇數(shù)節(jié)點(diǎn)的樹，且每次小動(dòng)作割出一個(gè)點(diǎn)對（包含Alice剛?cè)镜狞c(diǎn)），最后有兩種情況。

?- 出現(xiàn)某個(gè)結(jié)點(diǎn)有>=2個(gè)子節(jié)點(diǎn)為葉子節(jié)點(diǎn)。Alice染這個(gè)點(diǎn)，Bob跟不上Alice的節(jié)奏，出現(xiàn)孤點(diǎn)，Ailice取勝

?- 否則整個(gè)過程一定會(huì)持續(xù)到樹被染光或者Bob被Alice掏空導(dǎo)致做不了小動(dòng)作進(jìn)而被迫割出一塊size為奇數(shù)的子樹（這棵樹顯然沒辦法兩兩匹配）而敗北。

?- Bob被允許“任意時(shí)刻”做小動(dòng)作看似很厲害其實(shí)很雞肋，把問題改成“Bob只能在游戲開始之前做小動(dòng)作”會(huì)得到同樣的結(jié)論。 - “氪不改命，玄不救非” 時(shí)間復(fù)雜度 $O(n)$

我的理解：經(jīng)過長時(shí)間的思考確實(shí)，Bob的k次斷線必須維持所剩節(jié)點(diǎn)為偶，若為偶，那么白棋下在葉節(jié)點(diǎn)，使得再下的黑棋多染一個(gè)節(jié)點(diǎn)，使所剩棋子為奇數(shù)，為了不多染一個(gè)棋子，那么Bob要斷線避免所剩棋子數(shù)為奇數(shù)個(gè)。

Bob贏得方法只有一個(gè)，在開始之前先用他的k次斷線，將樹分成兩兩匹配，才能贏，否則Bob輸。

兩兩匹配方法，應(yīng)該是從葉節(jié)點(diǎn)上往上匹配，所以父節(jié)點(diǎn)必須在孩子節(jié)點(diǎn)匹配以后再匹配（拓?fù)渑判?#xff09;，沒有匹配的節(jié)點(diǎn)都應(yīng)該和他的父親匹配，那么只需要判斷他（他自己沒有染色，并且他的兒子都已經(jīng)匹配過）的父親是否被染色了即可，若未染色，則將他和他的父親都染上色，否則他沒法完成配對，則一定是白棋贏。

若兩兩可以匹配則判斷k次斷線是否夠，不夠則白贏；

我的AC代碼：

#include<stdio.h> #include <iostream> #include<string.h> #include<algorithm> #include<cmath> #include<map> #define LL long long using namespace std; int a[600];//a[i]=x表示節(jié)點(diǎn)i的父節(jié)點(diǎn)是x int b[600];//染色標(biāo)記 int topo[600];//排序后的節(jié)點(diǎn)順序 int c[600][600];//c[i][j]=x代表節(jié)點(diǎn)i的第j個(gè)兒子是節(jié)點(diǎn)x; int s[600];//s[i]=x代表節(jié)點(diǎn)i有x個(gè)兒子 int cnt; void toposort(int d) {for(int i=0;i<s[d];i++) toposort(c[d][i]);//先將d節(jié)點(diǎn)的兒子排序topo[cnt++]=d;//然后再放節(jié)點(diǎn)dreturn ; } int main() {int t;scanf("%d",&t);while(t--){memset(b,0,sizeof(b));memset(s,0,sizeof(s));cnt=0;b[0]=1;a[1]=0;//因?yàn)?節(jié)點(diǎn)沒有父節(jié)點(diǎn)，為了避免特判，所以給1添個(gè)已經(jīng)染色的父節(jié)點(diǎn)0；不影響結(jié)果。int n,k;scanf("%d%d",&n,&k);for(int i=2; i<=n; i++){scanf("%d",&a[i]);c[a[i]][s[a[i]]++]=i;}int flag=0;if(n%2) flag=1;//節(jié)點(diǎn)數(shù)為奇數(shù)一定是白贏if(!flag) toposort(1);if(!flag)for(int i=0; i<n; i++)if(!b[topo[i]]){if(b[a[topo[i]]]){flag=1;break;}else{b[a[topo[i]]]=1;b[topo[i]]=1;}}if(!flag) if(n/2-1>k) flag=1;//判斷k次夠不夠if(flag) printf("Alice\n");else printf("Bob\n");} }

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