HDU - 3516 Tree Construction
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HDU - 3516 Tree Construction
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HDU - 3516
思路:
平行四邊形不等式優(yōu)化dp
:)
代碼:
#pragma GCC optimize(2) #pragma GCC optimize(3) #pragma GCC optimize(4) #include<bits/stdc++.h> using namespace std; #define y1 y11 #define fi first #define se second #define pi acos(-1.0) #define LL long long #define LD long double //#define mp make_pair #define pb push_back #define ls rt<<1, l, m #define rs rt<<1|1, m+1, r #define ULL unsigned LL #define pll pair<LL, LL> #define pli pair<LL, int> #define pii pair<int, int> #define piii pair<int, pii> #define pdd pair<long double, long double> #define mem(a, b) memset(a, b, sizeof(a)) #define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0); //headconst int N = 1e3 + 5; int dp[N][N], s[N][N], x[N], y[N], n; int main() {while(~scanf("%d", &n)) {for (int i = 1; i <= n; ++i) scanf("%d %d", &x[i], &y[i]);for (int i = n; i >= 1; --i) {dp[i][i] = 0;s[i][i] = i;for (int j = i+1; j <= n; ++j) {dp[i][j] = 0x3f3f3f3f;for (int k = s[i][j-1]; k <= s[i+1][j]; ++k) {if(k+1 <= j && dp[i][k]+dp[k+1][j]+y[k]-y[j]+x[k+1]-x[i] < dp[i][j]) {dp[i][j] = dp[i][k]+dp[k+1][j]+y[k]-y[j]+x[k+1]-x[i];s[i][j] = k;}}}}printf("%d\n", dp[1][n]);}return 0; }?
轉(zhuǎn)載于:https://www.cnblogs.com/widsom/p/10955045.html
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