• CF607B Zuma(区间dp)


    题意

    题目链接

    Sol

    裸的区间dp,转移的时候判一下两个字符是否相等即可

    #include<bits/stdc++.h> 
    #define Pair pair<int, int>
    #define MP(x, y) make_pair(x, y)
    #define fi first
    #define se second
    #define LL long long 
    #define ull unsigned long long 
    #define Fin(x) {freopen(#x".in","r",stdin);}
    #define Fout(x) {freopen(#x".out","w",stdout);}
    using namespace std;
    const int MAXN = 5001, mod = 1e9 + 7, INF = 1e9 + 10;
    const double eps = 1e-9;
    template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
    template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
    template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
    template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
    template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
    template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
    template <typename A> inline void debug(A a){cout << a << '
    ';}
    template <typename A> inline LL sqr(A x){return 1ll * x * x;}
    inline int read() {
        char c = getchar(); int x = 0, f = 1;
        while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
        return x * f;
    }
    int N, a[MAXN], f[MAXN][MAXN];
    signed main() {
        N = read();
        for(int i = 1; i <= N; i++) a[i] = read(), f[i][i] = 1;
        for(int len = 2; len <= N; len++) {
            for(int l = 1; l + len - 1 <= N; l++) {
                int r = l + len - 1;
                f[l][r] = r - l + 1;
                for(int k = l; k < r; k++) chmin(f[l][r], f[l][k] + f[k + 1][r]);
                if(a[l] == a[r]) chmin(f[l][r], f[l + 1][r - 1] + (r - l == 1));
            }
        }
        cout << f[1][N];
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/zwfymqz/p/10488241.html
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