【DP】 POJ 1085 Triangle War

3935 단어 dppoj
상태는 2진법으로 압축해서 저장하고 기억화 검색하면 돼요...
#include <iostream>
#include <queue> 
#include <stack> 
#include <map> 
#include <set> 
#include <bitset> 
#include <cstdio> 
#include <algorithm> 
#include <cstring> 
#include <climits>
#include <cstdlib>
#include <cmath>
#include <time.h>
#define maxn 300005
#define maxm 400005
#define eps 1e-7
#define mod 1000000007
#define INF 0x3f3f3f3f
#define PI (acos(-1.0))
#define lowbit(x) (x&(-x))
#define mp make_pair
#define ls o<<1
#define rs o<<1 | 1
#define lson o<<1, L, mid 
#define rson o<<1 | 1, mid+1, R
#define pii pair<int, int>
#pragma comment(linker, "/STACK:16777216")
typedef long long LL;
typedef unsigned long long ULL;
//typedef int LL;
using namespace std;
LL qpow(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base;base=base*base;b/=2;}return res;}
LL powmod(LL a, LL b){LL res=1,base=a;while(b){if(b%2)res=res*base%mod;base=base*base%mod;b/=2;}return res;}
// head

int o = (1 << 18) - 1;
int dp[maxn][2];
int a[maxn];
int c[maxn];
int m, s, ss, res;

void init()
{
	a[0] = (1 << 0) | (1 << 1) | (1 << 2);
	a[1] = (1 << 3) | (1 << 4) | (1 << 7);
	a[2] = (1 << 4) | (1 << 5) | (1 << 2);
	a[3] = (1 << 5) | (1 << 6) | (1 << 8);
	a[4] = (1 << 9) | (1 << 10) | (1 << 15);
	a[5] = (1 << 7) | (1 << 10) | (1 << 11);
	a[6] = (1 << 11) | (1 << 12) | (1 << 16);
	a[7] = (1 << 12) | (1 << 13) | (1 << 8);
	a[8] = (1 << 13) | (1 << 14) | (1 << 17);
	a[9] = (1 << 0) | (1 << 3) | (1 << 1) | (1 << 6) | (1 << 7) | (1 << 8);
	for(int i = 0; i <= o; i++)
		for(int j = 0; j < 9; j++)
			if((i & a[j]) == a[j]) c[i]++;
	memset(dp, INF, sizeof dp);
}

void read()
{
	scanf("%d", &m);
	res = ss = 0;
	int t = 0, aa, bb, t2 = 0;
	for(int i = 0; i < m; i++) {
		scanf("%d%d", &aa, &bb);
		if(aa > bb) swap(aa, bb);
		if(aa == 1) {
			if(bb == 2) t |= (1 << 0);
			if(bb == 3) t |= (1 << 1);
		}
		if(aa == 2) {
			if(bb == 3) t |= (1 << 2);
			if(bb == 4) t |= (1 << 3);
			if(bb == 5) t |= (1 << 4);
		}
		if(aa == 3) {
			if(bb == 5) t |= (1 << 5);
			if(bb == 6) t |= (1 << 6);
		}
		if(aa == 4) {
			if(bb == 5) t |= (1 << 7);
			if(bb == 7) t |= (1 << 9);
			if(bb == 8) t |= (1 << 10);
		}
		if(aa == 5) {
			if(bb == 6) t |= (1 << 8);
			if(bb == 8) t |= (1 << 11);
			if(bb == 9) t |= (1 << 12);
		}
		if(aa == 6) {
			if(bb == 9) t |= (1 << 13);
			if(bb == 10) t |= (1 << 14);
		}
		if(aa == 7) t |= (1 << 15);
		if(aa == 8) t |= (1 << 16);
		if(aa == 9) t |= (1 << 17);
		if(!ss) res += c[t] - c[t2];
		else res -= c[t] - c[t2];
		if(!(c[t] - c[t2])) ss ^= 1;
		t2 = t;
	}
	s = t;
}

int dfs(int now, bool flag)
{
	if(dp[now][flag] != INF) return dp[now][flag];
	if(now == o) return dp[now][flag] = 0;
	int ans;
	if(flag) ans = -INF;
	else ans = INF;
	for(int i = 0; i < 18; i++) {
		if((now & (1 << i)) == (1 << i)) continue;
		int next = now | (1 << i);
		if(c[next] == c[now]) {
			if(flag) ans = max(ans, dfs(next, 0));
			else ans = min(ans, dfs(next, 1));
		}
		else {
			if(flag) ans = max(ans, dfs(next, 1) + c[next] - c[now]);
			else ans = min(ans, dfs(next, 0) - (c[next] - c[now]));
		}
	}
	return dp[now][flag] = ans;
}	

void work()
{
	res += dfs(s, ss ^ 1);
	if(res > 0) printf("A wins.
"); else printf("B wins.
"); } int main() { init(); int _, __ = 0; while(scanf("%d", &_)!=EOF) { __ = 0; while(_--) { read(); printf("Game %d: ", ++__); work(); } } return 0; }

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