Codeforces 811C Vladik and Memorable Trip[dp]
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题意:给了一串长度<=5000的序列,要求你选定一些区间,区间的值为其中不同的数的异或,而且所有相同的数,要不就不取,要不就要在一个区间内。
分析:我们可以先预处理出每个数第一次出现的位置st,最后出现的位置ed,还有区间的异或值sum
dp[i] 表示到了 i 之前的序列的答案为多少。
每次在ed的位置转移,判断其中是否有数的ed比现在大,如果大,跳过。没有的话,找其中最小的min_st,
dp[i] = min(dp[i-1],dp[min_st - 1] + sum[min_st][i]);
其他位置,dp[i] = dp[i - 1];
以下是代码:
int st[MAXN], n, ed[MAXN], vis[MAXN];ll sum[MAXN][MAXN], a[MAXN];ll dp[MAXN];int main() {memset(sum, 0, sizeof(sum));memset(vis, 0, sizeof(vis));memset(dp, 0, sizeof(dp));scanf("%d", &n);for (int i = 1; i <= n; i++) scanf("%I64d", &a[i]);for (int i = 1; i <= n; ++i) {if (!vis[a[i]])st[a[i]] = i;ed[a[i]] = i;vis[a[i]] = 1;}for (int i = 1; i <= n; ++i) {memset(vis, 0, sizeof(vis));for (int j = i; j <= n; ++j) {if (!vis[a[j]])sum[i][j] = sum[i][j - 1] ^ a[j];else sum[i][j] = sum[i][j - 1];vis[a[j]] = 1;}}dp[0] = 0;for (int i = 1; i <= n; ++i) {if (i == ed[a[i]]) {bool ok = 1;int L = st[a[i]];for (int j = st[a[i]]; j < ed[a[i]]; ++j) {if (ed[a[j]] > i) {ok = 0;break;}L = min(L, st[a[j]]);}if (ok) dp[i] = max(dp[i - 1], dp[L - 1] + sum[L][ed[a[i]]]);else dp[i] = dp[i - 1];}else dp[i] = dp[i - 1];}printf("%I64d\n", dp[n]);}
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