LightOJ 1051 - Good or Bad （dp）

题意:

给定一个字符串,连续3个元音字母或者连续5个辅音字母为BAD,如果完全不BAD为GOOD,"?"可以是任意字母,如果都有可能是MIXED

分析:

dp[i][j][k]:=第i个字母,连续j个元音或者连续k个辅音,的状态

转移就很蛋疼了−−反正最后ac了

代码:

//
//  Created by TaoSama on 2015-11-20
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n, dp[55][4][6]; //0Illegal 1GOOD 2BAD 3MIXED
char s[55];

bool isVowel(char c) {
return c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U';
}

void get(int&x, int y) {
if(x && x != y) x = 3;
else x = y;
}

//1GOOD 2BAD 3MIXED
int gao() {
bool good = false, bad = false;
for(int i = 1; i <= 3; ++i) {
if(!dp
[i][0]) continue;
if(dp
[i][0] == 1) good = true;
else if(dp
[i][0] == 2) bad = true;
else return 3;
}
for(int i = 1; i <= 5; ++i) {
if(!dp
[0][i]) continue;
if(dp
[0][i] == 1) good = true;
else if(dp
[0][i] == 2) bad = true;
else return 3;
}
if(good && bad) return 3;
else if(good) return 1;
else return 2;
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
int kase = 0;
while(t--) {
scanf("%s", s + 1);
n = strlen(s + 1);
memset(dp, 0, sizeof dp);
if(s[1] == '?' || isVowel(s[1])) dp[1][1][0] = 1;
if(s[1] == '?' || !isVowel(s[1])) dp[1][0][1] = 1;
for(int i = 1; i < n; ++i) {
for(int j = 1; j <= 3; ++j) {
if(!dp[i][j][0]) continue;
if(s[i + 1] == '?' || isVowel(s[i + 1])) {
if(j >= 2) dp[i + 1][3][0] = 2;
else get(dp[i + 1][j + 1][0], dp[i][j][0]);
}
if(s[i + 1] == '?' || !isVowel(s[i + 1]))
get(dp[i + 1][0][1], dp[i][j][0]);
}
for(int j = 1; j <= 5; ++j) {
if(!dp[i][0][j]) continue;
if(s[i + 1] == '?' || !isVowel(s[i + 1])) {
if(j >= 4) dp[i + 1][0][5] = 2;
else get(dp[i + 1][0][j + 1], dp[i][0][j]);
}
if(s[i + 1] == '?' || isVowel(s[i + 1]))
get(dp[i + 1][1][0], dp[i][0][j]);
}
}
int ans = gao();
string x = ans == 1 ? "GOOD" : ans == 2 ? "BAD" : "MIXED";
printf("Case %d: %s\n", ++kase, x.c_str());
}
return 0;
}

分析:

看了下大牛写的,dp[i][j][k]:=此状态是否可达,问题一下就简单了−−

代码:

//
//  Created by TaoSama on 2015-11-20
//  Copyright (c) 2015 TaoSama. All rights reserved.
//
//#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iomanip>
#include <iostream>
#include <map>
#include <queue>
#include <string>
#include <set>
#include <vector>

using namespace std;
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
const int N = 1e5 + 10, INF = 0x3f3f3f3f, MOD = 1e9 + 7;

int n;
bool dp[55][4][6];
char s[55];

bool isVowel(char c) {
return c == 'A' || c == 'E' || c == 'I' || c == 'O' || c == 'U';
}

int main() {
#ifdef LOCAL
freopen("C:\\Users\\TaoSama\\Desktop\\in.txt", "r", stdin);
//  freopen("C:\\Users\\TaoSama\\Desktop\\out.txt","w",stdout);
#endif
ios_base::sync_with_stdio(0);

int t; scanf("%d", &t);
int kase = 0;
while(t--) {
scanf("%s", s + 1);
n = strlen(s + 1);
memset(dp, false, sizeof dp);
dp[0][0][0] = true;
for(int i = 1; i <= n; ++i) {
for(int j = 0; j < 3; ++j) {
if(s[i] == '?' || isVowel(s[i]))
dp[i][j + 1][0] |= dp[i - 1][j][0];
if(s[i] == '?' || !isVowel(s[i]))
dp[i][0][1] |= dp[i - 1][j][0];
}
for(int j = 0; j < 5; ++j) {
if(s[i] == '?' || !isVowel(s[i]))
dp[i][0][j + 1] |= dp[i - 1][0][j];
if(s[i] == '?' || isVowel(s[i]))
dp[i][1][0] |= dp[i - 1][0][j];
}
}
bool good = false, bad = false;
for(int i = 1; i < 3; ++i) if(dp
[i][0]) good = true;
for(int i = 1; i < 5; ++i) if(dp
[0][i]) good = true;
for(int i = 1; i <= n; ++i) {
if(dp[i][3][0]) bad = true;
if(dp[i][0][5]) bad = true;
}
printf("Case %d: ", ++kase);
if(good && bad) puts("MIXED");
else if(good) puts("GOOD");
else puts("BAD");
}
return 0;
}