uva 11967 - Hic-Hac-Hoe
Q
— Hic-Hac-Hoe
Time Limit: 2 sec
Memory Limit: 32 MB
Everybody knows how to play tic-tac-toe. If accidentally you do not know the rules of this game, you can always consult Wikipedia.
In this problem we will use a slightly different version of tic-tac-toe. First, the game board is not limited to 3x3 cells, but considered infinite. Also in order to win a player must get not 3 but at least knoughts or crosses in a line (horizontal, vertical or diagonal).
In modified tic-tac-to version it is not so easy to determine a winner. So in this problem you will be given a list of turns performed by the players during the game and you need to determine the winner.
INPUT
There is a number of tests T (T ≤ 100) on the first line. Each test case is described by the two numbers n k (n ≤ 105, k ≤ 5), where n stands for number of turns for both players and k for winning line size. Next line contains n pairs of signed 32-bit integers x y — coordinates of each players turn. All turns have been performed sequentially by both players and crosses have always started a game.
OUTPUT
For each test case output a single line "Case T: S". Where T is the test case number (starting from 1) and Sis equal to "crosses" or "noughts" if one of them has a winning line. If nobody yet has won a game output"none" and if both players have winning lines output "error" for "S".
SAMPLE INPUT
2 3 2 0 0 1 1 1 0 4 2 0 0 -1 0 1 1 -1 1
SAMPLE OUTPUT
Case 1: crosses Case 2: errorProblem by: Aleksej Viktorchik; Leonid Sislo
Huge Easy Contest #2
水题一道，在无限大的棋盘上玩hic-hac-hoe，用map存，然后枚举每个点，最多八个方向拓展k步。
#include <cstdio>
#include <algorithm>
#include <vector>
#include <map>
#include <queue>
#include <iostream>
#include <stack>
#include <set>
#include <cstring>
#include <stdlib.h>
#include <cmath>
using namespace std;
typedef long long LL;
typedef pair<int, int> P;
const int maxn = 100 + 5;
int n, k;
map<P, int> M;
vector<P> v;
bool judge(int x, int y){
int kind = M[P(x, y)];
// right
int tag = 1;
for(int i = 1;i < k;i++){
int tx = x+i;
int ty = y;
if(M.count(P(tx, ty)) == 0 || M[P(tx, ty)] != kind){
tag = 0;
break;
}
}
if(tag) return true;
// left
tag = 1;
for(int i = 1;i < k;i++){
int tx = x-i;
int ty = y;
if(M.count(P(tx, ty)) == 0 || M[P(tx, ty)] != kind){
tag = 0;
break;
}
}
if(tag) return true;
//up
tag = 1;
for(int i = 1;i < k;i++){
int tx = x;
int ty = y+i;
if(M.count(P(tx, ty)) == 0 || M[P(tx, ty)] != kind){
tag = 0;
break;
}
}
if(tag) return true;
//down
tag = 1;
for(int i = 1;i < k;i++){
int tx = x;
int ty = y-i;
if(M.count(P(tx, ty)) == 0 || M[P(tx, ty)] != kind){
tag = 0;
break;
}
}
if(tag) return true;
//right up
tag = 1;
for(int i = 1;i < k;i++){
int tx = x+i;
int ty = y+i;
if(M.count(P(tx, ty)) == 0 || M[P(tx, ty)] != kind){
tag = 0;
break;
}
}
if(tag) return true;
//left up
tag = 1;
for(int i = 1;i < k;i++){
int tx = x-i;
int ty = y+i;
if(M.count(P(tx, ty)) == 0 || M[P(tx, ty)] != kind){
tag = 0;
break;
}
}
if(tag) return true;
//right down
tag = 1;
for(int i = 1;i < k;i++){
int tx = x+i;
int ty = y-i;
if(M.count(P(tx, ty)) == 0 || M[P(tx, ty)] != kind){
tag = 0;
break;
}
}
if(tag) return true;
//left down
tag = 1;
for(int i = 1;i < k;i++){
int tx = x-i;
int ty = y-i;
if(M.count(P(tx, ty)) == 0 || M[P(tx, ty)] != kind){
tag = 0;
break;
}
}
if(tag) return true;
return false;
}
int main(){
int t, kase = 0;
scanf("%d", &t);
while(t--){
kase++;
scanf("%d%d", &n, &k);
M.clear();
v.clear();
for(int i = 0;i < n;i++){
int x, y;
scanf("%d%d", &x, &y);
M[P(x, y)] = i%2;
v.push_back(P(x, y));
}
int tagc = 0, tagn = 0;
for(int i = 0;i < n;i++){
if(judge(v[i].first, v[i].second)){
if(i%2 == 0) tagc = 1;
else tagn = 1;
}
}
printf("Case %d: ", kase);
if(tagc == 1 && tagn == 1){
printf("error\n");
}
else if(tagc == 0 && tagn == 0){
printf("none\n");
}
else if(tagc == 1){
printf("crosses\n");
}
else{
printf("noughts\n");
}
}
return 0;
}