USACO Section 2.4

/*
ID: beihai2013
TASK: ttwo
LANG: C++
*/
/*
刚开始想复杂了，想通过记录数组然后求牛或者人经过此点的时间戳来求周期，然后求一个通项
后面发现这样的讨论不仅复杂，而且容易出现十分多的情况
然后仔细一想，假设是有人和牛周期的话，他们分别得周期不超过100，联合起来周期肯定不超过100 * 100
所以设一个周期上限值，如果还没遇到就退出就可以了
*/
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 10 + 2;
const int MAXM = 4 + 1;
//int dp[2][MAXN][MAXN][MAXM];
char g[MAXN][MAXN];
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, 1, 0, -1};
bool validPos(int x, int y){if(x >= 0 && x < 10 && y >= 0 && y < 10) return true; else return false;}
int x[2], y[2], dir[2];
int main()
{
freopen("ttwo.in", "r", stdin);
freopen("ttwo.out", "w", stdout);
int n = 10;
for(int i = 0 ; i < 10 ; i++) scanf("%s", g[i]);
for(int i = 0 ; i < 10 ; i++) {
for(int j = 0 ; j < 10 ; j++) {
if(g[i][j] == 'C') x[0] = i, y[0] = j;
else if(g[i][j] == 'F') x[1] = i, y[1] = j;
}
}
dir[0] = dir[1] = 0;
int res = 0;
int T = 100000 + 5;
int cnt = 0;
while(T--) {
if(x[0] == x[1] && y[0] == y[1]) {
res = cnt;
break;
}
for(int i = 0 ; i < 2 ; i++) {
int tx = dx[dir[i]] + x[i];
int ty = dy[dir[i]] + y[i];
if(validPos(tx, ty) && g[tx][ty] != '*') {
x[i] = tx, y[i] = ty;
}
else {
dir[i] = (dir[i] + 1) % 4;
}
}
cnt++;
}

printf("%d\n", res);
return 0;
}

/*
ID: beihai2013
TASK: maze1
LANG: C++
*/
/*
宽搜做迷宫，不过这次是一次朝一个方向跳两格
*/
#include <bits/stdc++.h>
using namespace std;
#define mp make_pair
#define pu push
#define fi first
#define se second
typedef pair<int,int> pii;
const int MAXN = 200 + 2;
char g[MAXN * 2 + 1][MAXN * 2 + 1];
int dp[MAXN * 2 + 1][MAXN * 2 + 1], n, m;
int dx[] = {-1, 0, 1, 0};
int dy[] = {0, -1, 0, 1};
queue<pii>que;
bool validPos(int x, int y){if(x >= 0 && x < n && y >= 0 && y < m) return true; else return false;}
int main()
{
freopen("maze1.in", "r", stdin);
freopen("maze1.out", "w", stdout);
while(scanf("%d%d", &m, &n) != EOF) {
n = n * 2 + 1;
m = m * 2 + 1;
getchar();
for(int i = 0 ; i < n ; i++) {
cin.getline(g[i], MAXN * 2 + 1);
}
memset(dp, -1, sizeof dp);
while(!que.empty()) que.pop();
for(int i = 0 ; i < n ; i++) {
if(g[i][0] == ' ') que.pu(mp(i, 1)), dp[i][1] = 1;
if(g[i][m - 1] == ' ') que.pu(mp(i, m - 2)), dp[i][m - 2] = 1;
}
for(int i = 0 ; i < m ; i++) {
if(g[0][i] == ' ') que.pu(mp(1, i)), dp[1][i] = 1;
if(g[n - 1][i] == ' ') que.pu(mp(n - 2, i)), dp[n - 2][i] = 1;
}

while(!que.empty()) {
pii p = que.front(); que.pop();
for(int i = 0 ; i < 4 ; i++) {
int tx1 = dx[i] + p.fi;
int ty1 = dy[i] + p.se;
if(validPos(tx1, ty1) && g[tx1][ty1] == ' ') {
int tx2 = dx[i] + tx1;
int ty2 = dy[i] + ty1;
if(validPos(tx2, ty2) && g[tx2][ty2] == ' ') {
if(dp[tx2][ty2] == -1) {
dp[tx2][ty2] = dp[p.fi][p.se] + 1;
que.pu(mp(tx2, ty2));
}
}
}
}
}

int res = -1;
for(int i = 0 ; i < n ; i++) {
for(int j = 0 ; j < m ; j++) {
res = max(res, dp[i][j]);
}
}

printf("%d\n", res);
}
return 0;
}

/*
ID: beihai2013
TASK: cowtour
LANG: C++
*/
/*
每个点存一个连通块，连通块持有信息为连通块包含的点、连通块直径
然后连接两个连通块时，假设连通块1通过点A连接连通块2的点B
得到的新答案为（连接A、B两点距离+连通块1内距A最远距离+连通块2内距B最远距离）
此新答案还要与原A、B的直径做比较
*/
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 150 + 5;
int n, vis[MAXN], g[MAXN][MAXN];
double x[MAXN], y[MAXN];
int bccno[MAXN], bcccnt;
char str[MAXN];
double dep[MAXN];
double r[MAXN], far[MAXN];
vector<int> bcc[MAXN];
queue<int> que;
double cal(double x1, double y1, double x2, double y2){return sqrt(1.0 * (x1 - x2) * (x1 - x2) + 1.0 * (y1 - y2) * (y1 - y2));}
void dfs(int u, int num)
{
bccno[u] = num;
bcc[num].push_back(u);
for(int i = 0 ; i < n ; i++) {
if(g[u][i] == 1 && bccno[i] == 0) {
dfs(i, num);
}
}
}
double bfs(int u)
{
double res = 0;
while(!que.empty()) que.pop();
for(int i = 0 ; i < (int)bcc[bccno[u]].size() ; i++) {
int v = bcc[bccno[u]][i];
dep[v] = -1.0, vis[v] = 0;
}
dep[u] = 0; vis[u] = 1;
que.push(u);
while(!que.empty()) {
int v = que.front(); que.pop(); vis[v] = 0;
//        if(u == 6) {
//            printf("v = %d, dep = %f\n", v, dep[v]);
//        }
for(int i = 0 ; i < n ; i++) {
if(g[v][i] == 0) continue;
double dis = cal(x[v], y[v], x[i], y[i]);
if(dep[i] > dep[v] + dis || dep[i] == -1.0) {
dep[i] = dep[v] + dis;
if(vis[i] == 0) {
vis[i] = 1;
que.push(i);
}
}
}
}
///此处需要单独写，之前在上面写WA了
for(int i = 0 ; i < (int)bcc[bccno[u]].size() ; i++) res = max(res, dep[bcc[bccno[u]][i]]);
//    if(u == 6)
//    system("pause");
return res;
}
double solve(int u, int v)
{
//    double t1 = bfs(u);
//    double t2 = bfs(v);
double t3 = cal(x[u], y[u], x[v], y[v]);
return max(r[bccno[u]], max(far[u] + far[v] + t3, r[bccno[v]]));
}
int main()
{
freopen("cowtour.in", "r", stdin);
freopen("cowtour.out", "w", stdout);
while(scanf("%d", &n) != EOF) {
for(int i = 0 ; i < n ; i++) scanf("%lf%lf", x + i, y + i);
for(int i = 0 ; i < n ; i++) {
scanf("%s", str);
for(int j = 0 ; j < n ; j++) {
if(str[j] == '0') g[i][j] = 0;
else g[i][j] = 1;
}
}

memset(bccno, 0, sizeof bccno);
bcccnt = 0;
for(int i = 0 ; i < n ; i++) {
if(bccno[i]) continue;
else {
bcc[++bcccnt].clear();
dfs(i, bcccnt);
}
}
memset(r, 0, sizeof r);
for(int i = 0 ; i < n ; i++) {
far[i] = bfs(i);
r[bccno[i]] = max(r[bccno[i]], far[i]);
}
double res = 1e5 * 1e5 * 1e5;
for(int i = 0 ; i < n ; i++) {
for(int j = i + 1 ; j < n ; j++) {
if(bccno[i] == bccno[j]) continue;
//                double temp = solve(i, j);
//                printf("i = %d, j = %d, temp = %f\n", i, j, temp);
//                system("pause");
res = min(res, solve(i, j));
}
}
printf("%.6f\n", res);
}
return 0;
}

/*
ID: beihai2013
TASK: comehome
LANG: C++
*/
/*
简单最短路
*/
#include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define fi first
#define se second
#define mp make_pair
typedef pair<int,int> pii;
const int MAXN = 100 + 5;
vector<pii>e[MAXN];
int toI(char c)
{
if(c >= 'A' && c <= 'Y') return c - 'A' + 1;
else if(c == 'Z') return 0;
else return c - 'a' + 26;
}
char s1[5], s2[5];
int dep[MAXN], vis[MAXN], m;
queue<int>que;
int main()
{
//    printf("toI(z) = %d, toI(Z) = %d, toI(Y) = %d, toI(a) = %d\n", toI('z'), toI('Z'), toI('Y'), toI('a'));
freopen("comehome.in", "r", stdin);
freopen("comehome.out", "w", stdout);
while(scanf("%d", &m) != EOF) {
for(int i = 0 ; i < MAXN ; i++) e[i].clear();
for(int i = 0 ; i < m ; i++) {
scanf("%s%s", s1, s2);
int w;  scanf("%d", &w);
int u = toI(s1[0]);
int v = toI(s2[0]);
e[u].pb(mp(v, w));
e[v].pb(mp(u, w));
}

for(int i = 0 ; i < MAXN ; i++) dep[i] = -1, vis[i] = 0;
while(!que.empty()) que.pop();
que.push(0), vis[0] = 1, dep[0] = 0;
while(!que.empty()) {
int u = que.front(); que.pop();
vis[u] = 0;
for(int i = 0 ; i < (int)e[u].size() ; i++) {
pii p = e[u][i];
if(dep[p.fi] == -1 || dep[p.fi] > dep[u] + p.se) {
dep[p.fi] = dep[u] + p.se;
if(vis[p.fi] == 0) {
vis[p.fi] = 1;
que.push(p.fi);
}
}
}
}

int res = -1;
int re = 1;
for(int i = 1 ; i < 26 ; i++) {
if(dep[i] != -1 && (res == -1 || res > dep[i]))
res = dep[i], re = i;
}
printf("%c %d\n", 'A' + re - 1, res);
}
return 0;
}

/*
ID: beihai2013
TASK: fracdec
LANG: C++
*/
/*
找循环节就可以，调格式调了一段时间
*/
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e6 + 5;
int dp[MAXN];
int out[MAXN], outcnt;
int pre;
char res[MAXN];
int main()
{
freopen("fracdec.in", "r", stdin);
freopen("fracdec.out", "w", stdout);
int n, d;
while(scanf("%d%d", &n, &d) != EOF) {
pre = n / d;
n = n % d;
n *= 10;
memset(dp, 0, sizeof dp);
int flag = 0;
outcnt = 0;
int rear = 1, head = 1;
while(1) {
//            printf("n = %d, outcnt = %d\n", n, outcnt);
//            printf("out = "); for(int i = 1 ; i <= outcnt ; i++) printf("%d", out[i]);
//            printf("\n");
//            system("pause");
if(n == 0) {
rear = outcnt;
head = outcnt + 1;
flag = 1;
break;
}
if(dp
) {
head = dp
;
rear = outcnt;
break;
}
else {
dp
= ++outcnt;
out[outcnt] = n / d;
n = n % d;
n = n * 10;
}
}
//        printf("head = %d, rear = %d, outcnt = %d, pre = %d, flag = %d\n", head, rear, outcnt, pre, flag);
int rescnt = 0;
if(pre == 0) res[rescnt++] = '0' + 0;
else {
while(pre) {
res[rescnt++] = pre % 10 + '0';
pre = pre / 10;
}
for(int i = 0 ; i < rescnt / 2 ; i++) swap(res[i], res[rescnt - 1 - i]);
}
res[rescnt++] = '.';
if(outcnt == 0) res[rescnt++] = '0';
for(int i = 1 ; i < head ; i++) {
res[rescnt++] = out[i] + '0';
}
if(flag == 0) {
res[rescnt++] = '(';
for(int i = head ; i <= rear ; i++) res[rescnt++] = out[i] + '0';
res[rescnt++] = ')';
}
for(int i = 0 ; i < rescnt ; i++) {
printf("%c", res[i]);
if(i % 76 == 75 && i != 0) printf("\n");
}
printf("\n");
}
return 0;
}