poj 1085 Triangle War (狀壓+記憶化搜索)
阿新 • • 發佈:2019-03-15
popu for align 技術 min \n most n) let
Triangle War
Description Triangle War is a two-player game played on the following triangular grid:?Two players, A and B, take turns filling in any dotted line connecting two dots, with A starting first. Once a line is filled, it cannot be filled again. If the line filled by a player completes one or more triangles, she owns the completed triangles and she is awarded another turn (i.e. the opponent skips a turn). The game ends after all dotted lines are filled in, and the player with the most triangles wins the game. The difference in the number of triangles owned by the two players is not important.? For example, if A fills in the line between 2 and 5 in the partial game on the left below:? Then, she owns the triangle labelled A and takes another turn to fill in the line between 3 and 5. B can now own 3 triangles (if he wishes) by filling in the line between 2 and 3, then the one between 5 and 6, and finally the one between 6 and 9. B would then make one more move before it is A‘s turn again.? Input Output For each game, print the game number and the result on one line as shown below. If A wins, print the sentence "A wins." If B wins, print "B wins."Sample Input 4 6 2 4 4 5 5 9 3 6 2 5 3 5 7 2 4 4 5 5 9 3 6 2 5 3 5 7 8 6 1 2 2 3 1 3 2 4 2 5 4 5 10 1 2 2 5 3 6 5 8 4 7 6 10 2 4 4 5 4 8 7 8 Sample Output Game 1: B wins. Game 2: A wins. Game 3: A wins. Game 4: B wins. Source field=source&key=East+Central+North+America+1999" rel="nofollow" style="text-decoration:none;">East Central North America 1999 |
題意:
兩個人玩遊戲,依次在三角形上放邊,假設能構成三角形。則獎勵繼續該此人放,問最後得到的三角形多。
思路:
給邊編號,記憶化搜索即可。做過好多這樣的題。就不多寫思路了。
代碼:
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define maxn 1005
#define MAXN 50005
#define mod 1000000009
#define INF 0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-6
typedef long long ll;
using namespace std;
int n,m,ans,cnt,tot,flag;
bool vis[10];
int dp[300000],mp[15][15],sc[5];
int tri[9][3]=
{
0,1,2,3,4,7,2,4,5,5,6,8,9,10,15,
7,10,11,11,12,16,8,12,13,13,14,17
};
int cal(int s)
{
int i,j,t=0;
for(j=0; j<9; j++)
{
if((s&(1<<tri[j][0]))&&(s&(1<<tri[j][1]))&&(s&(1<<tri[j][2]))) t++;
}
return t;
}
int dfs(int state,int score)
{
if(dp[state]!=-1) return dp[state];
int i,j,t,tst,num,best=0,tmp;
num=9-score;
for(i=0; i<=17; i++)
{
if(state&(1<<i)) continue ;
tst=state|(1<<i);
t=cal(tst);
if(t>num)
{
tmp=t-num+dfs(tst,score-(t-num));
best=max(best,tmp);
}
else
{
tmp=score-dfs(tst,score);
best=max(best,tmp);
}
}
dp[state]=best;
return best;
}
int main()
{
int i,j,t,test=0;
mp[1][2]=0;mp[1][3]=1;mp[2][3]=2;mp[2][4]=3;mp[2][5]=4;mp[3][5]=5;
mp[3][6]=6;mp[4][5]=7;mp[5][6]=8;mp[4][7]=9;mp[4][8]=10;mp[5][8]=11;
mp[5][9]=12;mp[6][9]=13;mp[6][10]=14;mp[7][8]=15;mp[8][9]=16;mp[9][10]=17;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
tot=0;
int x,y,z,turn=0,num=0;
sc[0]=sc[1]=0;
memset(vis,0,sizeof(vis));
for(i=1; i<=n; i++)
{
scanf("%d%d",&x,&y);
z=mp[x][y];
tot|=(1<<z);
flag=0;
for(j=0; j<9; j++)
{
if(vis[j]) continue ;
if((tot&(1<<tri[j][0]))&&(tot&(1<<tri[j][1]))&&(tot&(1<<tri[j][2])))
{
vis[j]=1;
num++;
flag=1;
sc[turn]++;
}
}
if(!flag) turn^=1;
}
memset(dp,-1,sizeof(dp));
z=dfs(tot,9-num);
sc[turn]+=z;
sc[turn^1]+=(9-num-z);
if(sc[0]>sc[1]) printf("Game %d: A wins.\n",++test);
else printf("Game %d: B wins.\n",++test);
}
return 0;
}
poj 1085 Triangle War (狀壓+記憶化搜索)