标签：mem amp space 数组 技术 check lap img its
二分+贪心+动态规划
第一问就是二分+贪心，和跳石头挺像的
第二问是dp，dp[i][j]表示第i次切割切到了第j段木棍，转移就是dp[i][j] = sigma(dp[i-1][k]), sum[j]-sum[k]<=ans,ans是最大长度，这里第j段木棍表示现在正在分割1-j这些木棍。很明显这样时间空间都不满足，空间滚动数组就行了，时间我们需要用类似单调队列优化一下，其实也是双指针。每次sum[j]-sum[q[l]]不满足就向前移动
初值是dp[0][i]=(sum[i] <= ans)
最终答案是sigma(dp[0->m][n])
#include<bits/stdc++.h> using namespace std; const int N = 50010, mod = 10007; int n, m; int l[N], dp[2][N], sum[N], q[N]; bool check(int mid) { int tot = 0, ret = 0; for(int i = 1; i <= n; ++i) if(l[i] > mid) return false; for(int i = 1; i <= n; ++i) { if(tot + l[i] > mid) tot = l[i], ++ret; else tot += l[i]; } return ret <= m; } int main() { scanf("%d%d", &n, &m); for(int i = 1; i <= n; ++i) scanf("%d", &l[i]), sum[i] = sum[i - 1] + l[i]; int l = 0, r = sum[n] + 1, ans = 0; while(r - l > 1) { int mid = (l + r) >> 1; if(check(mid)) r = ans = mid; else l = mid; } int pre = 0, ret = 0; printf("%d ", ans); // dp[pre][0] = 1; for(int i = 1; i <= n; ++i) dp[pre][i] = (sum[i] <= ans); for(int i = 1; i <= m; ++i) { pre ^= 1; // memset(dp[pre], 0, sizeof(dp[pre])); int l = 1, r = 0, tot = 0; for(int j = 1; j <= n; ++j) { while(sum[j] - sum[q[l]] > ans && l <= r) tot = ((tot - dp[pre ^ 1][q[l]]) % mod + mod) % mod, ++l; dp[pre][j] = tot; tot = ((tot + dp[pre ^ 1][j]) % mod + mod) % mod; q[++r] = j; } ret = (ret + dp[pre][n]) % mod; } printf("%d\n", ret); return 0; }
标签：mem amp space 数组 技术 check lap img its
原文地址：http://www.cnblogs.com/19992147orz/p/7462646.html
