洛谷P4170 [CQOI2007]涂色(区间dp)

题意

题目链接

sol

震惊，某知名竞赛网站竟照搬省选原题！

裸的区间dp，\(f[l][r]\)表示干掉\([l, r]\)的最小花费，昨天写的时候比较困于是就把能想到的转移都写了。。

// luogu-judger-enable-o2
// luogu-judger-enable-o2
#include<bits/stdc++.h> 
#define pair pair<int, int>
#define mp(x, y) make_pair(x, y)
#define fi first
#define se second
#define int long long 
#define ll long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 1e6 + 10, 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 << '\n';}
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;
char s[maxn];
int f[501][501];
signed main() {
    scanf("%s", s + 1);
    n = strlen(s + 1);
    memset(f, 0x3f, sizeof(f));
    for(int i = 1; i <= n; i++) 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;
            if(s[l] == s[l + 1]) chmin(f[l][r], f[l + 1][r]);
            else chmin(f[l][r], f[l + 1][r] + 1);
            if(s[r] == s[r - 1]) chmin(f[l][r], f[l][r - 1]);
            else chmin(f[l][r], f[l][r - 1] + 1);
            if(s[l] == s[l + 1] && s[r] == s[r - 1]) chmin(f[l][r], min(min(f[l + 1][r], f[l][r - 1]), f[l + 1][r - 1] + 1));
            else chmin(f[l][r], f[l + 1][r - 1] + 2);
            if(s[l] == s[r]) chmin(f[l][r],min(min(f[l + 1][r], f[l][r - 1]), f[l + 1][r - 1] + 1));
            for(int k = l; k < r; k++) chmin(f[l][r], f[l][k] + f[k + 1][r]);
        }
    }
    cout << f[1][n];
    return 0;
}