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  • [USACO08FEB]修路Making the Grade

    题目链接:

    走这里

    题目分析:

    考虑绝对值的几何意义,显然(b)里的数一定在(a)里出现过
    离不离散化问题不大,用下标作第二位状态就行
    (dp[i][j])表示第(i)个数,高度为(a[j])时的最优解
    方程见代码

    代码:

    #include<bits/stdc++.h>
    #define int long long
    #define N (2000 + 10)
    using namespace std;
    inline int read() {
    	int cnt = 0, f = 1; char c = getchar();
    	while (!isdigit(c)) {if (c == '-') f = -f; c = getchar();}
    	while (isdigit(c)) {cnt = (cnt << 3) + (cnt << 1) + c - '0'; c = getchar();}
    	return cnt * f;
    }
    int n, a[N], b[N << 1], dp[N][N], gmin = 1 << 30;
    int ans = (1 << 30);
    signed main() {
    //	freopen("grading.in", "r", stdin);
    //	freopen("grading.out", "w", stdout);
    	n = read();
    	for (register int i = 1; i <= n; ++i) a[i] = b[i] = read();
    	memset(dp, 0x3f, sizeof(dp));
    	for (register int i = 1; i <= n; ++i) dp[1][i] = abs(a[1] - a[i]);
    	sort(a + 1, a + n + 1);
    	
    	for (register int i = 2; i <= n; ++i) {
    		gmin = (1 << 30);
    		for (register int j = 1; j <= n; ++j) {
    			gmin = min(gmin, dp[i - 1][j]);
    			dp[i][j] = gmin + abs(b[i] - a[j]);
    		}
    	}
    	for (register int i = 1; i <= n; ++i) ans = min(ans, dp[n][i]);
    
    	memset(dp, 0x3f, sizeof(dp));
    	
    	for (register int i = 2; i <= n; ++i) {
    		gmin = (1 << 30);
    		for (register int j = n; j >= 1; --j) {
    			gmin = min(gmin, dp[i - 1][j]);
    			dp[i][j] = min(dp[i][j], gmin + abs(b[i] - a[j]));
    		}
    	}
    	
    	for (register int i = 1; i <= n; ++i) ans = min(ans, dp[n][i]);
    	printf("%lld", ans);
    	return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/kma093/p/11620160.html
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