标签：map 最大 ret change ted http target add site
problem1 link
令$f[x1][y1][x2][y2]$表示矩形(x1,y1)(x2,y2)中能选出的最大值。dp即可。
problem2 link
这个题应该有更好的递推公式。
我的做法是这样的。设$f[i]$表示$i$时的答案，令$g[i]=f[i]-f[i-1]$。通过暴力计算$g$的前几项，发现$g$的公式为
$g(n+2)=a(n)=\frac{(2n+3)*(6n^4+36n^3+76n^2+66n+5)}{960}-\frac{(-1)^n}{64}$
problem3 link
由于对于一个节点，其对于某一种tree只能选择至多一次，因此可以对每种tree依次进行处理。
对于tree的某一种，找到target中与其同构的所有子树，然后进行dp即可。 设$f[mask]$将$mask$对应的边反转的最小代价。
code for problem1
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
public class DonutsOnTheGridEasy {
public int calc(String[] grid) {
final int n = grid.length;
final int m = grid[0].length();
if (n < 3 || m < 3) {
return 0;
}
int[][] h = new int[n + 1][m + 1];
int[][] v = new int[n + 1][m + 1];
for (int i = 0; i < n; ++i) {
for (int j = 0; j < m; ++j) {
h[i + 1][j + 1] = h[i + 1][j];
if (grid[i].charAt(j) == ‘0‘) {
h[i + 1][j + 1] += 1;
}
}
}
for (int j = 0; j < m; ++j) {
for (int i = 0; i < n; ++i) {
v[i + 1][j + 1] = v[i][j + 1];
if (grid[i].charAt(j) == ‘0‘) {
v[i + 1][j + 1] += 1;
}
}
}
int[][][][] f = new int[n + 1][m + 1][n + 1][m + 1];
for (int height = 3; height <= n; ++ height) {
for (int width = 3; width <= m; ++ width) {
for (int i = 1; i + height -1 <= n; ++ i) {
for (int j = 1; j + width -1 <= m; ++ j) {
int i1 = i + height - 1;
int j1 = j + width -1;
f[i][j][i1][j1]= Math.max(f[i][j][i1][j1], f[i + 1][j][i1][j1]);
f[i][j][i1][j1]= Math.max(f[i][j][i1][j1], f[i][j][i1 - 1][j1]);
f[i][j][i1][j1]= Math.max(f[i][j][i1][j1], f[i][j + 1][i1][j1]);
f[i][j][i1][j1]= Math.max(f[i][j][i1][j1], f[i][j][i1][j1 - 1]);
if (!check(i, j, i1, j1, h, v)) {
continue;
}
f[i][j][i1][j1]= Math.max(f[i][j][i1][j1], f[i + 1][j + 1][i1 - 1][j1 - 1] + 1);
}
}
}
}
return f[1][1][n][m];
}
boolean check(int x1, int y1, int x2, int y2, int[][] h, int[][] v) {
return h[x1][y2] - h[x1][y1 - 1] == y2 - y1 + 1
&& h[x2][y2] - h[x2][y1 - 1] == y2 - y1 + 1
&& v[x2][y1] - v[x1 - 1][y1] == x2 - x1 + 1
&& v[x2][y2] - v[x1 - 1][y2] == x2 - x1 + 1;
}
}
code for problem2
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
public class ConvexHexagons {
final static int mod = 1000000007;
final static long rev960 = pow(960, mod - 2);
final static long rev64 = pow(64, mod - 2);
public int find(int N) {
if (N < 3) {
return 0;
}
int ans = 0;
for (int i = 3; i <= N; ++i) {
ans += cal(i - 2);
ans %= mod;
}
return ans;
}
static long pow(long a, long b) {
a %= mod;
long ans = 1;
while (b > 0) {
if (1 == (b & 1)) {
ans = ans * a % mod;
}
a = a * a % mod;
b >>= 1;
}
return ans;
}
long cal(long n) {
long A = (2 * n + 3) % mod;
long B = (6 * pow(n, 4) + 36 * pow(n, 3) + 76 * pow(n, 2) + 66 * n + 5) % mod;
long ans = A * B % mod * rev960 % mod;
if (1 == (n & 1)) {
ans += rev64;
}
else {
ans -= rev64;
}
ans %= mod;
if (ans < 0) {
ans += mod;
}
return ans;
}
}
code for problem3
import java.util.*;
public class ActivateTree {
List<Edge> alledges = null;
List<Edge> currentTree = null;
List<List<Integer>> mask = new ArrayList<>();
boolean[] visited = null;
public int minCost(String[] trees, String target, int[] costs) {
alledges = getEdges(target);
int n = 0;
for (int i = 0; i < alledges.size(); ++ i) {
n = Math.max(n, alledges.get(i).x + 1);
n = Math.max(n, alledges.get(i).y + 1);
}
for (int i = 0; i < n; ++ i) {
mask.add(new ArrayList<>());
}
final int m = alledges.size();
int[][] f = new int[2][1 << m];
Arrays.fill(f[0], -1);
visited = new boolean[m];
f[0][0] = 0;
for (int i = 0; i < trees.length; ++ i) {
currentTree = getEdges(trees[i]);
for (int j = 0; j < n; ++ j) {
mask.get(j).clear();
}
dfs(0, new int[currentTree.size()]);
for (int u = 0; u < n; ++ u) {
for (int j = 0; j < (1 << m); ++ j) {
f[1][j] = f[0][j];
}
for (int s = 0; s < (1 << m); ++ s) {
for (int t: mask.get(u)) {
if (f[1][s ^ t] != -1) {
if (f[0][s] == -1 || f[0][s] > f[1][s ^ t] + costs[i]) {
f[0][s] = f[1][s ^ t] + costs[i];
}
}
}
}
}
}
return f[0][(1 << m) - 1];
}
Map<Integer,Integer> map0 = new HashMap<>();
Map<Integer,Integer> map1 = new HashMap<>();
void dfs(int id, int[] ch) {
if (id == currentTree.size()) {
int st = 0;
for (int i : ch) {
st |= 1 << i;
}
int rt = alledges.get(ch[0]).x;
while (true) {
boolean tag = false;
for (int i = 0; i < ch.length; ++ i) {
if (alledges.get(ch[i]).y == rt) {
rt = alledges.get(ch[i]).x;
tag = true;
break;
}
}
if (!tag) {
break;
}
}
if (mask.get(rt).contains(st)) {
return;
}
map0.clear();
map1.clear();
for (int i = 0; i < ch.length; ++ i) {
int x = alledges.get(ch[i]).x;
int y = alledges.get(ch[i]).y;
int xx = currentTree.get(i).x;
int yy = currentTree.get(i).y;
for (int j = 0; j < 2; ++ j) {
int u = (j == 0)? x : y;
int v = (j == 0)? xx : yy;
if (map0.containsKey(u)) {
if (map0.get(u) != v) {
return;
}
if (!map1.containsKey(v) || map1.get(v) != u) {
return;
}
}
else {
if (map1.containsKey(v)) {
return;
}
map0.put(u, v);
map1.put(v, u);
}
}
}
mask.get(rt).add(st);
return;
}
for (int i = 0; i < alledges.size(); ++ i) {
if (!visited[i]) {
visited[i] = true;
ch[id] = i;
dfs(id + 1, ch);
visited[i] = false;
}
}
}
static List<Edge> getEdges(String target) {
List<Edge> result = new ArrayList<>();
String[] all = target.split("\\s+");
for (int i = 1; i < all.length; ++ i) {
int p = Integer.valueOf(all[i]);
result.add(new Edge(p, i));
}
return result;
}
static class Edge {
int x, y;
Edge() {}
Edge(int x, int y) {
this.x = x;
this.y = y;
}
}
}
标签：map 最大 ret change ted http target add site
原文地址：http://www.cnblogs.com/jianglangcaijin/p/7751371.html
