luogu [TJOI2007]线段

题目链接

luogu [TJOI2007]线段

题解

dp[i][0/1]第i行在左/右端点的最短路
瞎转移

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 = 500007;  
int n; 
int l[maxn],r[maxn],dp[maxn][2]; 
int main() {  
    n = read(); 
    for(int i = 1;i <= n;++ i)   l[i] = read(),r[i] = read(); 
    dp[1][0] = r[1] + r[1] - l[1] - 1; 
    dp[1][1] = r[1] - 1; 
    for(int i = 2;i <= n;++ i) { 
    	int x = dp[i - 1][0],y = dp[i - 1][1]; 
       	dp[i][0] = min(x + abs(l[i-1] - r[i]) + r[i] - l[i] + 1, y + abs(r[i-1] - r[i]) + r[i] - l[i] + 1); 
       	dp[i][1] = min(x + abs(l[i-1] - l[i]) + r[i] - l[i] + 1, y + abs(r[i-1] - l[i]) + r[i] - l[i] + 1); 
    }   
    printf("%d
",min(dp[n][0] + n - l[n], dp[n][1] + n - r[n])); 
    return 0; 
}