1.Luogu 4170 [CQOI2007]涂色
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define rep(i, a, b) for (int i = a; i <= b; ++i)
4 const int N = 57;
5
6 int n, f[N][N]; char a[N];
7
8 int main() {
9 scanf("%s", a + 1); n = strlen(a + 1); memset(f, 0x3f, sizeof(f));
10 rep(i, 1, n) {
11 f[i][i] = 1;
12 }
13
14 rep(len, 2, n) rep(i, 1, n - len + 1) { int j = i + len - 1;
15 rep(k, i, j - 1) {
16 if (a[i] == a[j]) f[i][j] = min(f[i][j], f[i][k] + f[k + 1][j] - 1);
17 else f[i][j] = min(f[i][j], f[i][k] + f[k + 1][j]);
18 }
19 }
20
21 printf("%d\n", f[1][n]);
22 return 0;
23 }
2.Luogu 1063 能量项链
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define rep(i, a, b) for (int i = a; i <= b; ++i)
4 const int N = 207;
5
6 int n, a[N], f[N][N], ans;
7
8 int main() {
9 scanf("%d", &n);
10 rep(i, 1, n) {
11 scanf("%d", &a[i]);
12 a[n + i] = a[i];
13 }
14
15 rep(len, 2, n + 1) rep(i, 1, n * 2 - len + 1) { int j = i + len - 1;
16 rep(k, i + 1, j - 1) {
17 f[i][j] = max(f[i][j], f[i][k] + f[k][j] + a[i] * a[k] * a[j]);
18 }
19 // cout << i << ‘ ‘ << j << ‘ ‘ << f[i][j] << ‘\n‘;
20 if (len == n + 1) ans = max(ans, f[i][j]);
21 }
22
23 printf("%d\n", ans);
24 return 0;
25 }
3.Luogu 1005 矩阵取数游戏
1 #include <bits/stdc++.h>
2 using namespace std;
3 typedef long long ll;
4 #define rep(i, a, b) for (ll i = a; i <= b; ++i)
5 const ll N = 87;
6
7 ll n, m, a[N], f[N][N], ans;
8
9 int main() {
10 scanf("%lld%lld", &n, &m);
11
12 while (n--) {
13 memset(f, 0, sizeof(f));
14 rep(i, 1, m) {
15 scanf("%lld", &a[i]);
16 f[i][i] = a[i] * (1 << m);
17 }
18 rep(len, 2, m) rep(i, 1, m - len + 1) { ll j = i + len - 1;
19 f[i][j] = max(f[i][j], max(f[i][j - 1] + a[j] * (1 << (m + 1 - len)), f[i + 1][j] + a[i] * (1 << (m + 1 - len))));
20 }
21 ans += f[1][m];
22 }
23 printf("%lld\n", ans);
24 return 0;
25 }
4.Luogu 1220 关路灯
f[i][j][state] : [i, j]区间 state=0 当前选i state = 1 当前选j
注意枚举的顺序
转移的设计时 在同时刻不在[i, j]区间里的数也要考虑 不符合局部最优性
1 #include <bits/stdc++.h>
2 using namespace std;
3 #define rep(i, a, b) for (int i = a; i <= b; ++i)
4
5 const int N = 57;
6
7 int n, c, p[N], v[N], s[N], f[N][N][2];
8
9 inline int cal(int x, int y, int l, int r) {
10 return (p[y] - p[x]) * (s[n] - s[r] + s[l - 1]);
11 }
12
13 int main() {
14 scanf("%d%d", &n, &c);
15
16 memset(f, 0x3f, sizeof(f));
17 f[c][c][0] = f[c][c][1] = 0;
18
19 rep(i, 1, n) {
20 scanf("%d%d", &p[i], &v[i]);
21 s[i] = s[i - 1] + v[i];
22 }
23
24 rep(l, 2, n) rep(i, 1, n - l + 1) {
25 int j = i + l - 1;
26 f[i][j][0] = min(f[i + 1][j][0] + cal(i, i + 1, i + 1, j), f[i + 1][j][1] + cal(i, j, i + 1, j));
27 f[i][j][1] = min(f[i][j - 1][0] + cal(i, j, i, j - 1), f[i][j - 1][1] + cal(j - 1, j, i, j - 1));
28 }
29
30 printf("%d\n", min(f[1][n][0], f[1][n][1]));
31
32 return 0;
33 }
原文:https://www.cnblogs.com/Fo0o0ol/p/11319444.html