UVALive 6042 Bee Tower（DP）

题目链接：https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4053

题意：一个蜜蜂要从地面跳上最高塔，每次跳的高度不能超过H，而且只能跳到相邻的塔上，它每次最多可以将某个塔移动W个距离，花费为高度差 * 移动距离，问它要跳上最高塔所需的最小花费

思路：dp[i][j]代表把第i个塔移动到j位置所需的最小花费，预处理出那些最高塔，对于每个最高塔考虑蜜蜂跳回地面的花费，这样会有左右跳两种选择，逐个塔转移即可，到某个塔时如果此塔的高度 <= H，则更新答案，注意一开始就可以从最高塔跳下来的情况

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <vector>
#include <utility>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <climits>
#include <functional>
#include <deque>
#include <ctime>

#define lson l, mid, rt << 1
#define rson mid + 1, r, rt << 1 | 1
#pragma comment(linker, "/STACK:102400000,102400000")

using namespace std;

typedef long long ll;

const int maxn = 100100;
const int inf = 0x3f3f3f3f;

int dp[110][610];
int p[110], h[110];

int n, H, W;
vector <int> v;

int solvel(int s)
{
if (h[s] <= H) return 0;
memset(dp, inf, sizeof(dp));
dp[s][p[s]] = 0;

int res = inf;
for (int i = s - 1; i >= 1; i--)
{
if (h[i + 1] - h[i] > H) return res;

for (int j = p[s]; j >= p[1]; j--)
{
if (dp[i + 1][j] == inf) continue;

for (int k = j - W; k <= j - 1; k++)
{
dp[i][k] = min(dp[i][k], dp[i + 1][j] + abs(k - p[i]) * h[i]);
if (h[i] <= H)
res = min(res, dp[i][k]);
}
}
}
return res;
}

int solver(int s)
{
if (h[s] <= H) return 0;
memset(dp, inf, sizeof(dp));
dp[s][p[s]] = 0;

int res = inf;
for (int i = s + 1; i <= n; i++)
{
if (h[i - 1] - h[i] > H) return res;

for (int j = p[s]; j <= p
; j++)
{
if (dp[i - 1][j] == inf) continue;

for (int k = j + 1; k <= j + W; k++)
{
dp[i][k] = min(dp[i][k], dp[i - 1][j] + abs(k - p[i]) * h[i]);
if (h[i] <= H)
res = min(res, dp[i][k]);
}
}
}
return res;
}

int main()
{ int t;
scanf("%d", &t);
for (int ca = 1; ca <= t; ca++)
{
printf("Case #%d: ", ca);

int ma = -1;
cin >> n >> H >> W;
for (int i = 1; i <= n; i++)
{
scanf("%d%d", &p[i], &h[i]);
ma = max(ma, h[i]);
}

v.clear();
for (int i = 1; i <= n; i++)
if (h[i] == ma)
v.push_back(i);

// for(int i = 0; i < v.size(); i++)
// printf("%d ", v[i]);

int ans = inf;
for (int i = 0; i < v.size(); i++)
{
ans = min(solvel(v[i]), ans);
ans = min(solver(v[i]), ans);
}
if (ans == inf) ans = -1;
cout << ans << endl;
}
return 0;
}