HDU - 3516 Tree Construction(四边形优化)

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题目大意：给你一系列的坐标，x递增，y递减，问如何连成一棵树，能使数边和达到最小

解题思路：用dp[i][j]表示从第i到第j个连接起来的最小和
则dp[i][j] = dp[i][k] + dp[k + 1][j] + abs(x[k + 1] - x[i] + y[k] - y[j])

#include <cstdio>#include <cstring>#define abs(x)((x)>0?(x):(-(x)))const int N = 1010;const int INF = 0x3f3f3f3f;int x[N], y[N];int dp[N][N], s[N][N];int n;void init() {    for (int i = 1; i <= n; i++) {        scanf("%d%d", &x[i], &y[i]);        dp[i][i] = 0;        s[i][i] = i;    }}void solve() {    for (int len = 1; len < n; len++) {        for (int i = 1; i + len <= n; i++) {            int j = i + len;            dp[i][j] = INF;            for (int k = s[i][j - 1]; k <= s[i + 1][j]; k++)                if (dp[i][j] > dp[i][k] + dp[k + 1][j] + abs(x[k + 1] - x[i] + y[k] - y[j])) {                    dp[i][j] = dp[i][k] + dp[k + 1][j] + abs(x[k + 1] - x[i] + y[k] - y[j]);                    s[i][j] = k;                }        }    }    printf("%d\n", dp[1][n]);}int main() {    while (scanf("%d", &n) != EOF) {        init();        solve();    }    return 0;}

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