bzoj 1055 [HAOI2008]玩具取名 区间dp

题面

题目传送门

解法

直接区间dp即可

时间复杂度：\(O(16n^3)\)

代码

#include <bits/stdc++.h>
#define N 210
using namespace std;
struct Node {
    int x, y;
} a[5][N];
int s[5], p[5][5][5];
bool f[N][N][5], vis[N][N][5];
int num(char ch) {
    if (ch == ‘W‘) return 1;
    if (ch == ‘I‘) return 2;
    if (ch == ‘N‘) return 3;
    return 4;
}
bool dp(int l, int r, int k) {
    if (r - l + 1 <= 2) return f[l][r][k];
    if (vis[l][r][k]) return f[l][r][k];
    vis[l][r][k] = 1;
    for (int i = l; i < r; i++)
        for (int j = 1; j <= s[k]; j++)
            f[l][r][k] |= (dp(l, i, a[k][j].x) & dp(i + 1, r, a[k][j].y));
    return f[l][r][k];
}
int main() {
    for (int i = 1; i <= 4; i++) cin >> s[i];
    for (int i = 1; i <= 4; i++)
        for (int j = 1; j <= s[i]; j++) {
            string st; cin >> st;
            p[num(st[0])][num(st[1])][i] = 1;
            a[i][j] = (Node) {num(st[0]), num(st[1])};
        }
    string st; cin >> st;
    int len = st.size() - 1;
    for (int i = 0; i <= len; i++) f[i][i][num(st[i])] = 1;
    for (int i = 0; i < len; i++)
        for (int j = 1; j <= 4; j++)
            f[i][i + 1][j] = p[num(st[i])][num(st[i + 1])][j];
    bool flag = false;
    for (int i = 1; i <= 4; i++) {
        f[0][len][i] = dp(0, len, i);
        if (!f[0][len][i]) continue;
        flag = true;
        if (i == 1) cout << ‘W‘;
        if (i == 2) cout << ‘I‘;
        if (i == 3) cout << ‘N‘;
        if (i == 4) cout << ‘G‘;
    }
    if (!flag) cout << "The name is wrong!\n";
    return 0;
}

原文地址：https://www.cnblogs.com/copperoxide/p/9476727.html