题意: 有一根长度为l的木棍,木棍上面有m个切割点,每一次切割都要付出当前木棍长度的代价,问怎样切割有最小代价。
分析: 石子合并的逆过程。状态:设F(i,j)为区间( i, j )内最少的消耗是多少,则有转移方程:F(i,j)= min(F(i,k)+F(k,j)+cost(i,j)) {其中 i < k < j };这里注意,dp过程按区间大小递增顺序进行。
代码:
#include<bits/stdc++.h>
using namespace std;
#define INF 100000000
const int maxn = 60;
int dp[maxn][maxn];
int num[maxn];
int main() {
int n,L,q;
int MIN;
while(~scanf("%d",&L) && L) {
scanf("%d",&n);
memset(dp,0,sizeof(dp));
for(int i = 1; i <= n ; i++)
scanf("%d",&num[i]);
num[0] = 0;
num[n + 1] = L;
for(int i = 1; i <= n + 1; i++) {/// 枚举区间长度
for(int j = 0; j <= n + 1; j++) { /// 枚举区间起点
q = i + j ; /// 终点
MIN = INF;
if(q > n + 1) break;
for(int k = j + 1; k < q; k++) { ///枚举起点中间的分区间点
int temp = dp[j][k] + dp[k][q] + num[q] - num[j];
if(MIN > temp) MIN = temp;
}
if(MIN != INF) dp[j][q] = MIN;
}
}
printf("The minimum cutting is %d.\n",dp[0][n + 1]);
}
return 0;
}
刘汝佳代码(记忆化搜索):
#include<cstdio>
#include<cstring>
using namespace std;
const int maxn = 50 + 5;
int n, L, a[maxn], vis[maxn][maxn], d[maxn][maxn];
int dp(int i, int j)
{
if(i >= j - 1) return 0;
if(vis[i][j]) return d[i][j];
vis[i][j] = 1;
int& ans = d[i][j];
ans = -1;
for(int k = i+1; k <= j-1; k++)
{
int v = dp(i,k) + dp(k,j) + a[j] - a[i];
if(ans < 0 || v < ans) ans = v;
}
return ans;
}
int main()
{
while(scanf("%d%d", &L, &n) == 2 && L)
{
for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
a[0] = 0;
a[n+1] = L;
memset(vis, 0, sizeof(vis));
printf("The minimum cutting is %d.\n", dp(0, n+1));
}
return 0;
}