P4170 [CQOI2007]涂色 题解

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由数据范围易知，本题为区间 DP。
设状态 $f_{i,j}$ 表示 $i$ ~ $j$ 区间完成涂色时所需的最少的涂色次数
则有转移方程：$f_{i,j}=max\{f_{i,k}+f_{k+1,j}\}(k\in [i,j))$ 初值：$\mathrm{\forall }i\in [1,n]f_{i,i}=1$（区间长度为 1 时显然最少涂色次数为 1） 目标：$f_{1,n}$
上区间 DP 模板

#include <bits/stdc++.h>
using namespace std;


//    0. Enough array size? Enough array size? Enough array size? Integer overflow?

//    1. Think TWICE, Code ONCE!
//    Are there any counterexamples to your algo?

//    2. Be careful about the BOUNDARIES!
//    N=1? P=1? Something about 0?

//    3. Do not make STUPID MISTAKES!
//    Time complexity? Memory usage? Precision error?


#define debug(args...) fprintf(stderr, ##args)
#define gc getchar
using ll = long long;

const ll inf = 1e18;
const int N = 57;

int n, f[N][N];
char s[N];

int 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 = 1; len < n; ++len)
        for (int l = 1, r = l + len; r <= n; ++l, ++r)
            for (int k = l; k < r; ++k) f[l][r] = min(f[l][r], f[l][k] + f[k + 1][r]);

    printf("%d", f[1][n]);

    return 0;
}

结果
显然，我们刚才没有考虑特殊情况

特殊情况

易知当输入的字符串转移时若 $s_l$ == $s_r$ 为特殊情况，我们需要特判这种情况。
此时便不必对 $l,r$ 区间染色。选择对区间 $l,r-1$ 染色，或对区间 $l+1,r$ 染色，两者中取次小值即可。
状态转移方程变为：$f_{i,j}=min\{f_{i,j-1},f_{i+1,j}\}(\mathrm{\forall }{s}_i=s_j)$
至此，本题就可以正常的通过了
code

#include <bits/stdc++.h>
using namespace std;


//    0. Enough array size? Enough array size? Enough array size? Integer overflow?

//    1. Think TWICE, Code ONCE!
//    Are there any counterexamples to your algo?

//    2. Be careful about the BOUNDARIES!
//    N=1? P=1? Something about 0?

//    3. Do not make STUPID MISTAKES!
//    Time complexity? Memory usage? Precision error?


#define debug(args...) fprintf(stderr, ##args)
#define gc getchar
using ll = long long;

const ll inf = 1e18;
const int N = 57;

int n, f[N][N];
char s[N];

int 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 = 1; len < n; ++len)
        for (int l = 1, r = l + len; r <= n; ++l, ++r)
            if (s[l] == s[r]) f[l][r] = min(f[l][r - 1], f[l + 1][r]);
            else for (int k = l; k < r; ++k) f[l][r] = min(f[l][r], f[l][k] + f[k + 1][r]);
    printf("%d", f[1][n]);
    return 0;
}

AC