标签：期望 raw 部分 character bre update 4.0 oid max
problem1 link
首先预处理在已选字母的状态为$state$时是否可达。
然后就是按照题目进行dp。设$f[i]$表示已选字母集合为$i$时的结果。
每次可以根据$i$中含有的字母是奇数还是偶数个来确定现在该轮到谁选择。
problem2 link
交点可以分为三部分：
（1）已经确定的线之间的交点；
（2）未确定的线之间的交点；
（3）已确定的线与未确定线之间的交点。
第一部分比较容易计算。
对于第二部分，设$f[i]$表示$i$对点之间连线交点期望。那么对于$f[i+1]$来说，就是在最后添加一对点，然后枚举这一对点中的一个与对面集合中那个点相连，可以得到：$f[i+1]=\frac{1}{i+1}\sum_{k=1}^{i+1}(f[i]+k-1)=f[i]+\frac{i-1}{2}$,进而可以直接计算出$f[n]$的通项公式。
对于第三部分，可以枚举下半部分每个未匹配的点与每条确定的点有交点的概率。
problem3 link
code for problem1
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
public class DoorsGame {
static final int N = 1 << 16;
int[] bitNum = new int[N];
boolean[] g1 = null;
boolean[] g2 = null;
int[] f = new int[N];
int validMask = 0;
public int determineOutcome(String doors, int trophy) {
for (int i = 1; i < N; ++ i) {
bitNum[i] = bitNum[i >> 1] + (i & 1);
}
String s1 = doors.substring(0, trophy);
String s2 = (new StringBuilder(doors.substring(trophy))).reverse().toString();
g1 = cal(s1);
g2 = cal(s2);
Arrays.fill(f, Integer.MAX_VALUE);
for (int i = 0; i < doors.length(); ++ i) {
validMask |= 1 << (doors.charAt(i) - ‘A‘);
}
return dfs(0);
}
int dfs(int mask) {
if (g1[mask]) {
if (g2[mask]) {
return 0;
}
return bitNum[mask];
}
else {
if (g2[mask]) {
return -bitNum[mask];
}
}
if (f[mask] != Integer.MAX_VALUE) {
return f[mask];
}
if (contains(bitNum[mask], 0)) {
for (int i = 0; i < 16; ++ i) {
if (!contains(mask, i) && contains(validMask, i)) {
int ans = dfs(mask | (1 << i));
if (f[mask] == Integer.MAX_VALUE) {
f[mask] = ans;
continue;
}
if (ans < 0) {
if (f[mask] >= 0 || f[mask] < ans) {
f[mask] = ans;
}
}
else if (ans == 0) {
if (f[mask] > 0) {
f[mask] = ans;
}
}
else {
if (f[mask] > 0 && f[mask] < ans) {
f[mask] = ans;
}
}
}
}
}
else {
for (int i = 0; i < 16; ++ i) {
if (!contains(mask, i)) {
int ans = dfs(mask | (1 << i));
if (f[mask] == Integer.MAX_VALUE) {
f[mask] = ans;
continue;
}
if (ans < 0) {
if (f[mask] < 0 && f[mask] > ans) {
f[mask] = ans;
}
}
else if (ans == 0) {
if (f[mask] < 0) {
f[mask] = ans;
}
}
else {
if (f[mask] <= 0 || f[mask] > ans) {
f[mask] = ans;
}
}
}
}
}
return f[mask];
}
boolean[] cal(String s) {
boolean[] result = new boolean[N];
for (int i = 0; i < N; ++ i) {
int p = 0;
while (p < s.length()) {
if (contains(i, s.charAt(p) - ‘A‘)) {
p += 1;
}
else {
break;
}
}
if (p == s.length()) {
result[i] = true;
}
}
return result;
}
boolean contains(int mask, int t) {
return (mask & (1 << t)) != 0;
}
}
code for problem2
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
public class DrawingLines {
public double countLineCrossings(int n, int[] startDot, int[] endDot) {
double result = 0;
final int m = startDot.length;
for (int i = 0; i < m; ++ i) {
for (int j = i + 1; j < m; ++ j) {
boolean upper = startDot[i] < startDot[j];
boolean lower = endDot[i] < endDot[j];
if (upper != lower) {
result += 1;
}
}
}
result += (n - m) * (n - m - 1) / 4.0;
boolean[] b = new boolean[n + 1];
int[] pSum = new int[n + 1];
for (int i = 0; i < m; ++ i) {
b[endDot[i]] = true;
pSum[startDot[i]] = 1;
}
for (int i = 1; i <= n; ++ i) {
pSum[i] += pSum[i - 1];
}
if (n == m) {
return result;
}
for (int i = 1; i <= n; ++ i) {
if (b[i]) {
continue;
}
for (int j = 0; j < m; ++ j) {
if (endDot[j] < i) {
result += 1.0 * (startDot[j] - pSum[startDot[j]]) / (n - m);
}
else {
result += 1.0 * (n - startDot[j] - (m - pSum[startDot[j]])) / (n - m);
}
}
}
return result;
}
}
code for problem3
import java.util.*;
import java.math.*;
import static java.lang.Math.*;
public class BuildingRoads {
int getCost(char c) {
if (‘a‘ <= c && c <= ‘z‘) {
return c - ‘a‘ + 1;
}
if (‘A‘ <= c && c <= ‘Z‘) {
return (c - ‘A‘ + 1) * 100;
}
if (‘1‘ <= c && c <= ‘9‘) {
return (c - ‘0‘) * 10000;
}
if (c == ‘0‘) {
return 100000;
}
return 0;
}
Map<Character,Integer> map = new HashMap<>();
int getIndex(Character c) {
if (c == ‘!‘ || c == ‘@‘ || c == ‘#‘ || c == ‘$‘) {
if (map.containsKey(c)) {
return map.get(c) + 1;
}
int s = map.size();
map.put(c, s * 2);
return s * 2;
}
return -1;
}
int[][] g = null;
int n,m;
int[][][] f = null;
List<Integer> listx = new ArrayList<>();
List<Integer> listy = new ArrayList<>();
public int destroyRocks(String[] field) {
n = field.length;
m = field[0].length();
g = new int[n][m];
for (int i = 0; i < n; ++ i) {
for (int j = 0; j < m; ++ j) {
g[i][j] = getCost(field[i].charAt(j));
int id = getIndex(field[i].charAt(j));
if (id != -1) {
while (listx.size() < id + 1) {
listx.add(0);
listy.add(0);
}
listx.set(id, i);
listy.set(id, j);
}
}
}
final int M = listx.size();
f = new int[1 << M][n][m];
for (int i = 0; i < (1 << M); ++ i) {
for (int j = 0; j < n; ++ j) {
Arrays.fill(f[i][j], -1);
}
}
for (int i = 0; i < M; ++ i) {
f[1 << i][listx.get(i)][listy.get(i)] = 0;
}
for (int i = 0; i < (1 << M); ++ i) {
update(i);
}
int[] dp = new int[1 << (M / 2)];
Arrays.fill(dp, -1);
for (int i = 0; i < dp.length; ++ i) {
int mask = 0;
for (int j = 0; j < M / 2; ++ j) {
if (contains(i, 1 << j)) {
mask |= 3 << (j * 2);
}
}
for (int x = 0; x < n; ++ x) {
for (int y = 0; y < m; ++ y) {
int t = f[mask][x][y] + g[x][y];
if (t != -1 && (dp[i] == -1 || dp[i] > t)) {
dp[i] = t;
}
}
}
}
for (int i = 0; i < dp.length; ++ i) {
for (int j = 0; j < i; ++ j) {
if (!contains(i, j)) {
continue;
}
if (dp[j] != -1 && dp[i ^ j] != -1) {
int t = dp[j] + dp[i ^ j];
if (dp[i] == -1 || dp[i] > t) {
dp[i] = t;
}
}
}
}
return dp[dp.length - 1];
}
static class Node {
int cost;
int x;
int y;
Node() {}
Node(int cost, int x, int y) {
this.cost = cost;
this.x = x;
this.y = y;
}
}
public static class NodeCompare implements Comparator<Node> {
@Override
public int compare(Node c1, Node c2) {
return c1.cost - c2.cost;
}
};
Queue<Node> queue = new PriorityQueue<>(new NodeCompare());
void upd(int x, int y, int cost, int mask) {
if (f[mask][x][y] == -1 || f[mask][x][y] > cost) {
f[mask][x][y] = cost;
queue.offer(new Node(cost, x, y));
}
}
boolean contains(int x, int y) {
return (x & y) == y;
}
int[] dx = {1, -1, 0, 0};
int[] dy = {0, 0, 1, -1};
void update(int mask) {
for (int i = 0; i < mask; ++ i) {
if (!contains(mask, i)) {
continue;
}
for (int x = 0; x < n; ++ x) {
for (int y = 0; y < m; ++ y) {
if (f[i][x][y] != -1 && f[mask ^ i][x][y] != -1) {
int t = f[i][x][y] + f[mask ^ i][x][y];
if (f[mask][x][y] == -1 || f[mask][x][y] > t) {
f[mask][x][y] = t;
}
}
}
}
}
while (!queue.isEmpty()) {
queue.poll();
}
for (int x = 0; x < n; ++ x) {
for (int y = 0; y < m; ++ y) {
if (f[mask][x][y] != -1) {
queue.add(new Node(f[mask][x][y], x, y));
}
}
}
while (!queue.isEmpty()) {
Node node = queue.poll();
int cost = node.cost;
int x = node.x;
int y = node.y;
if (f[mask][x][y] != cost) {
continue;
}
for (int i = 0; i < 4; ++ i) {
int xx = x + dx[i];
int yy = y + dy[i];
if (xx < 0 || xx >= n || yy < 0 || yy >= m) {
continue;
}
int newCost =cost;
if (g[x][y] != g[xx][yy]) {
newCost += g[x][y];
}
upd(xx, yy, newCost, mask);
}
}
}
}
标签：期望 raw 部分 character bre update 4.0 oid max
原文地址：http://www.cnblogs.com/jianglangcaijin/p/7789094.html
