noip模拟赛 遭遇

分析：暴力挺好打的，对于前30%的数据神搜，hi相同的数据将所有的建筑按照c从小到大排序，看最多能跳多少,ci=0的数据将所有的建筑按照h从小到大排序，枚举起点和终点，看能否跳这么多,取个max就可以了.这样70分就到手了.

      部分分的提示还是比较明显的，要消除一个参数的影响，那么就按照h从小到大排序，显然只有可能顺着跳过城市，不能跳过去又跳回来.那么就是一个比较简单的dp了：f[i][j]表示跳了i次，最后一次跳到j的最小花费，转移的话枚举j之前的k就能转移了，最后倒叙枚举i看哪一个f[i][j]<=T就可以了.

      多个参数有影响的常见策略是消除一个参数的影响，常见的办法就是排序,如果一个点不能经过多次，那么就想一个办法让它强行不经过这个点.

暴力：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;int n, T, ans, vis[1010];
bool flag1 = true, flag2 = true;struct node
{int c, h;
}e[55];bool cmp1(node a, node b)
{return a.c < b.c;
}bool cmp2(node a, node b)
{return a.h < b.h;
}void dfs(int u, int sum,int tot)
{ans = max(ans, tot + 1);vis[u] = 1;for (int i = 1; i <= n; i++){if (!vis[i]){int huafei = abs(e[i].h - e[u].h) + e[u].c;if (sum + huafei <= T)dfs(i, sum + huafei, tot + 1);}}
}int main()
{scanf("%d", &n);for (int i = 1; i <= n; i++){scanf("%d", &e[i].c);if (e[i].c != 0)flag2 = false;}for (int i = 1; i <= n; i++){scanf("%d", &e[i].h);if (i != 1 && e[i].h != e[i - 1].h)flag1 = false;}scanf("%d", &T);if (n <= 5 || (!flag1 && !flag2)){for (int i = 1; i <= n; i++){memset(vis, 0, sizeof(vis));dfs(i,0,0);}printf("%d\n", ans);}elseif (flag1){sort(e + 1, e + 1 + n, cmp1);int res = 0, cur = 1;while (cur <= n && res <= T){ans++;res += e[cur].c;cur++;}printf("%d\n", ans - 1);}else{sort(e + 1, e + 1 + n, cmp2);for (int i = 1; i <= n; i++){int j = i + 1, res = 0;while (j <= n && res <= T){res += abs(e[j].h - e[j-1].h);j++;}ans = max(ans, j - i + 1);}printf("%d\n", ans - 1);}return 0;
}

正解：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <iostream>
#include <algorithm>using namespace std;int n,T;
int f[60][60];struct node
{int c, h;
}e[60];bool cmp(node a, node b)
{return a.h < b.h;
}int main()
{memset(f, 127 / 3, sizeof(f));scanf("%d", &n);for (int i = 1; i <= n; i++)scanf("%d", &e[i].c);for (int i = 1; i <= n; i++)scanf("%d", &e[i].h);scanf("%d", &T);sort(e + 1, e + 1 + n, cmp);f[0][1] = e[1].c;for (int i = 0; i <= n; i++)for (int j = 1; j <= n; j++)for (int k = j + 1; k <= n; k++)f[i + 1][k] = min(f[i + 1][k], f[i][j] + e[k].h - e[j].h + e[k].c);for (int i = n; i >= 0; i--)for (int j = 1; j <= n; j++)if (f[i][j] <= T){printf("%d\n", i + 1);return 0;}printf("0\n");return 0;
}