题意:给定n,m,有n个站点,m个炸弹能炸掉m段铁路,使得最后剩下的价值最小,计算方法看题目中的计算公式。
分析:以前用斜率优化写过一次,这次用四边形优化再写一次。我们设dp[i][j]表示前i段铁路炸j个炸弹的最小价值,那么显然有dp[i][j]=min(dp[k][j-1]+w[k+1][i]),w[i]][j]表示i~j是一个区间的价值,显然w[i][j]=w[i][j-1]+a[j]*(sum[j-1]-sum[i-1])。很容易证明w是满足区间包含单调性和四边形不等式的,那么用四边形优化就是了。O(n^2)
代码:
#include<map>
#include<set>
#include<cmath>
#include<queue>
#include<bitset>
#include<math.h>
#include<vector>
#include<string>
#include<stdio.h>
#include<cstring>
#include<iostream>
#include<algorithm>
#pragma comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int N=1010;
const int mod=100000000;
const int MOD1=1000000007;
const int MOD2=1000000009;
const double EPS=0.00000001;
typedef long long ll;
const ll MOD=1000000007;
const int INF=1000000010;
const ll MAX=1000000000000;
const double pi=acos(-1.0);
typedef double db;
typedef unsigned long long ull;
int a[N],s[N][N];
ll sum[N],w[N][N],dp[N][N];
int main()
{
int i,j,k,n,m;
while (scanf("%d%d", &n, &m)&&!(n==0&&m==0)) {
for (i=1;i<=n;i++) {
scanf("%d", &a[i]);sum[i]=sum[i-1]+a[i];
}
for (i=1;i<=n;i++)
for (j=i;j<=n;j++) w[i][j]=w[i][j-1]+a[j]*(sum[j-1]-sum[i-1]);
for (i=1;i<=n;i++)
for (j=1;j<=m;j++) dp[i][j]=MAX;
for (i=1;i<=n;i++) dp[i][0]=w[1][i],s[i][0]=0;
for (j=1;j<=m;j++) {
s[n+1][j]=n;
for (i=n;i>j;i--)
for (k=s[i][j-1];k<=s[i+1][j];k++)
if (dp[k][j-1]+w[k+1][i]<=dp[i][j]) {
dp[i][j]=dp[k][j-1]+w[k+1][i];s[i][j]=k;
}
}
printf("%I64d\n", dp[n][m]);
}
return 0;
}