codeforces Simba on the Circle (dp)

Simba on the Circle

 题意：长度为n的环，第i个数字为ai。 从起点s出发，走一个格子的花费为1，输出当前数字的花费为0，输出n个数的非递减序列的最小花费是多少，并打印方案。

若ai都不相同的话就很好做了，考虑值相同的ai作为一段，其中肯定有一个起点和终点，即这一段第一个走的i和最后一个走的i，从起点向同一个方向走到终点走完所有ai肯定是最优的。（终点就在起点旁边→_→）

dp1[i]表示以i为起点，走值相同的ai，的答案

dp2[i]表示以i为终点

转移：

dp2[i] = min(dp1[j] + dis[i][j]),  aj为比ai大的最小的一个

如果ai相同的只有一个，dp1[i] = dp2[i]

否则, dp1[i] = min(dp2[j] + cd[i][j]),  ai = aj注意方向大概要走一圈的样子

边界 dp2[i] = 0 (ai = max)

#include <iostream>#include <string>#include <cstring>#include <cstdio>#include <algorithm>#include <cmath>#include <vector>#pragma comment(linker, "/STACK:102400000,102400000")using namespace std;#define LL long long#define N 2010#define M 200020#define inf (1100000000)#define ls (i << 1)#define rs (ls | 1)#define md (ll + rr >> 1)#define lson ll, md, ls#define rson md + 1, rr, rsint d[N][N], cd[N][N], dp1[N], dp2[N];int nxt[N], a[N];int n, s;int gao2(int u) ;int gao1(int u);int gao2(int u) {int &res = dp2[u];if(res != -1) return res;if(nxt[u] == inf) return res = 0;res = inf;for(int i = 0; i < n; ++i) {if(a[i] == nxt[u]) {res = min(res, gao1(i) + d[u][i]);}}//printf("dp2[%d] %d\n", u, dp2[u]);return res;}int gao1(int u) {int &res = dp1[u];if(res != -1) return res;res = inf;for(int dir = -1; dir <= 1; dir += 2) {int v = -1;for(int i = 1; i < n; ++i) {int t = (u + dir * i + n ) % n;if(a[t] == a[u]) {v = t;break;}}if(v == -1)res = min(res, gao2(u));else {int add = dir == -1 ? cd[u][v] : cd[v][u];res = min(res, gao2(v) + add);}}//printf("dp1[%d] %d\n", u, dp1[u]);return res;}void output1(int u);void output2(int u);void output2(int u){int res = inf;if(nxt[u] == inf) return ;for(int i = 0; i < n; ++i) {if(a[i] == nxt[u]) {res = min(res, dp1[i] + d[u][i]);}if(res == dp2[u]) {printf("%c%d\n", (i-u+n)%n == d[u][i] ? '+' : '-', d[u][i]);output1(i);return ;}}}void output1(int u) {int res = inf;for(int dir = -1; dir <= 1; dir += 2) {int v = -1;for(int i = 1; i < n; ++i) {int t = (u + dir * i + n ) % n;if(a[t] == a[u]) {v = t;break;}}if(v == -1)res = min(res, dp2[u]);else {int add = dir == -1 ? cd[u][v] : cd[v][u];res = min(res, dp2[v] + add);}if(res == dp1[u]) {int last = 0;dir = (dir == -1 ? 1 : -1);for(int i = 1; i < n; ++i) {int t = (u + dir*i + n) % n;if(a[t] == a[u] || t == v) {printf("%c%d\n", dir == -1 ? '-' : '+', i-last);last = i;}}if(v == -1) output2(u);else output2(v);return ;}}}int main() { //freopen("in.txt", "r", stdin);scanf("%d%d", &n, &s);--s;int mi = inf;for(int i = 0; i < n; ++i)scanf("%d", &a[i]), mi = min(mi, a[i]);for(int i = 0; i < n; ++i) {nxt[i] = inf;for(int j = 0; j < n; ++j) {if(a[j] > a[i])nxt[i] = min(nxt[i], a[j]);}}for(int i = 0; i < n; ++i) {for(int j = 0; j < n; ++j) {d[i][j] = min(abs(i-j), n-abs(i-j));cd[i][j] = j >= i ? j - i : n - (i-j);}}memset(dp1, -1, sizeof dp1);memset(dp2, -1, sizeof dp2);int ans = inf;for(int i = 0; i < n; ++i) {if(a[i] == mi)ans = min(ans, gao1(i) + d[i][s]);}printf("%d\n", ans);int res = inf;for(int i = 0; i < n; ++i) {if(a[i] == mi)  {res = min(res, dp1[i] + d[i][s]);if(res == ans) {printf("%c%d\n", (i-s+n)%n == d[s][i] ? '+' : '-', d[s][i]);output1(i);return 0;}}}}