• Bzoj 1055 玩具取名(区间DP)


    题面

    题解

    字符很麻烦,不妨用数字代替(比如1代表'W')

    const char c[5] = {0, 'W', 'I', 'N', 'G'};
    

    接着,像这种两个子串可以合并成另一个子串的题可以考虑区间$DP$

    设$bool$数组$f_{i,j,k}$表示区间$[l,r]$能否合成单个字符$c_k$

    于是就可以套区间$DP$的板子了

    #include <map>
    #include <cstdio>
    #include <cstring>
    #include <algorithm>
    using std::min; using std::max;
    using std::sort; using std::swap;
    using std::unique; using std::lower_bound;
    using std::map;
    typedef long long ll;
    
    template<typename T>
    void read(T &x) {
        int flag = 1; x = 0; char ch = getchar();
        while(ch < '0' || ch > '9') { if(ch == '-') flag = -flag; ch = getchar(); }
        while(ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar(); x *= flag;
    }
    
    const int Len = 2e2 + 10, _ = 5;
    const char c[5] = {0, 'W', 'I', 'N', 'G'};
    char s[Len]; bool p, f[Len][Len][_];
    int num[_], list[Len][_], tot, len;
    int check(char x) { for(int i = 1; i <= 4; ++i) if(x == c[i]) return i; }
    
    int main () {
    	for(int i = 1; i <= 4; ++i) read(num[i]);
    	for(int i = 1; i <= 4; ++i)
    		for(int j = 1; j <= num[i]; ++j) {
    			char ss[5]; scanf("%s", ss);
    			list[++tot][0] = i;
    			list[tot][1] = check(ss[0]);
    			list[tot][2] = check(ss[1]);
    		}
    	scanf("%s", s + 1), len = strlen(s + 1);
    	for(int i = 1; i <= len; ++i) f[i][i][check(s[i])] = true;
    	for(int k = 1; k <= len; ++k)
    		for(int i = 1; i + k <= len; ++i) {
    			int j = i + k;
    			for(int p = i; p < j; ++p)
    				for(int l = 1; l <= tot; ++l)
    					if(f[i][p][list[l][1]] && f[p + 1][j][list[l][2]])
    						f[i][j][list[l][0]] = true;
    		}
    	for(int i = 1; i <= 4; ++i)
    		if(f[1][len][i]) putchar(c[i]), p = true;
    	if(!p) printf("The name is wrong!");
    	puts("");
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/water-mi/p/10150621.html
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