luoguP4360 [CEOI2004]锯木厂选址

题目链接

luoguP4360 [CEOI2004]锯木厂选址

题解

dis:后缀和

sum:前缀和

补集转化，减去少走的，得到转移方程

dp[i] = min(tot - sumj * disj - (sumi - sumj) * disi

不需要斜率优化吧？反正也是个SB式子

代码

#include<bits/stdc++.h>
using namespace std;
inline int read() {
    int x = 0,f = 1;
    char c = getchar();
    while(c < '0' ||c > '9')c = getchar();
    while(c <= '9' && c >= '0')x = x* 10 + c -'0',c = getchar();
    return x *f ;
}
const int maxn  = 200007;
int dp[maxn],dis[maxn],sum[maxn];
int tot = 0;
int main() {
    int n = read();
    for(int a,b,i = 1;i <= n;++ i) {
        a = read(),b = read();
        sum[i] = sum[i - 1] + a; dis[i] += b;
        tot += sum[i] * b;
    }
    for(int i = n;i >= 1;-- i) dis[i] = dis[i] + dis[i + 1];
    //printf("%d\n",tot);
    memset(dp,0x3f,sizeof dp);
    int ans = 0x3f3f3f3f;
    for(int i = 1;i <= n;++ i) {
        for(int j = 1;j < i;++ j)  {
            dp[i] = min(tot - sum[j] * dis[j] - (sum[i] - sum[j]) * dis[i],dp[i]);
        }
        ans = min(ans,dp[i]);
    }
    printf("%d\n",ans);
    return 0;
}