【google apec 2015 1c】 挖矿、打地铁 最短路djstra、+*计算器
Problem A. Minesweeper
http://code.google.com/codejam/contest/5214486/dashboard#s=p0Problem B. Taking Metro
http://code.google.com/codejam/contest/5214486/dashboard#s=p1Problem C. Broken Calculator
http://code.google.com/codejam/contest/5214486/dashboard#s=p2Problem D. Tetris
http://code.google.com/codejam/contest/5214486/dashboard#s=p3第一题：贪心法，先消除0
#include <iostream>
using namespace std;
typedef long long int ll;
#define MOD 1000000007
char cc[302][302];
int tt[302][302];
int dr[8][2]={{-1,-1},{-1,1},{1,1},{1,-1},{0,1},{0,-1},{1,0},{-1,0}};
inline int getNum(int j,int k,int N){
int cnt=0;
for(int i=0;i<8;i++){
int x=j+dr[i][0],y=k+dr[i][1];
if(x>=0&&x<N&&y>=0&&y<N)
cnt+=cc[x][y]=='*'?1:0;
}
return cnt;
}
void dfs(int j,int k,int N){
int tmp=tt[j][k];
tt[j][k]=-1;
if(tmp!=0) return;
for(int i=0;i<8;i++){
int x=j+dr[i][0],y=k+dr[i][1];
if(x>=0&&x<N&&y>=0&&y<N){
dfs(x,y,N);
}
}
}
int main(){
int n;cin>>n;
for(int i=1;i<=n;i++){
int N;cin>>N;
for(int j=0;j<N;j++){
for(int k=0;k<N;k++){
char c;cin>>c;
cc[j][k]=c;
}
}
for(int j=0;j<N;j++){
for(int k=0;k<N;k++){
// *=-2,
char c=cc[j][k];
tt[j][k]=c=='*'?-2:getNum(j,k,N);
}
}
/*
for(int j=0;j<N;j++){
for(int k=0;k<N;k++){
cout<<tt[j][k]<<" ";
}cout<<endl;
}
*/
int cnt=0;
for(int j=0;j<N;j++){
for(int k=0;k<N;k++){
if(tt[j][k]==0){
cnt++;
dfs(j,k,N);
}
}
}
/*
for(int j=0;j<N;j++){
for(int k=0;k<N;k++){
cout<<tt[j][k]<<" ";
}cout<<endl;
}
*/
for(int j=0;j<N;j++){
for(int k=0;k<N;k++){
if(tt[j][k]!=-1&&tt[j][k]!=-2){
cnt++;
}
}
}
cout<<"Case #"<<i<<": "<<cnt<<endl;
}
return 0;
}
第二题：最短路径（dilijstra、floyd？）
注意起点车站，可以换乘换到别的路线上的车站。
但是没A，不知道为什么。。。
#include<cstdio>
#include <assert.h>
#include<cstring>
#include<cmath>
#include <vector>
#include <iostream>
using namespace std;
int time_[1001][1001];
int d[1001][1001];
int wait[1001];
int start[1001];
void floyd(int n){
for(int j=0;j<n;j++){
for(int k=0;k<n;k++){
d[j][k]=time_[j][k];
}
}
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
for(int k=0;k<n;k++){
if(d[j][i]>=0&&d[i][k]>=0){
if(d[j][k]<0||d[j][k]>d[j][i]+d[i][k])
d[j][k]=d[j][i]+d[i][k];
}
}
}
}
}
int dkl(int time_[][1001],int l,int r,int n){
bool visit[1001];
int d[1001];
memset(d,-1,sizeof(d));
memset(visit,0,sizeof(visit));
d[l]=0;
int s=l;
while(s!=r){
visit[s]=true;
int minT=99999999,minI=-1;
for(int i=0;i<n;i++){
if(!visit[i]){
if(time_[s][i]>=0&&(d[i]<0||d[s]+time_[s][i]<d[i])){
d[i]=d[s]+time_[s][i];
//cout<<"go to "<<i<<" from "<<s<<" with "<<d[s]<<" to "<<d[i]<<endl;
}
if(d[i]>=0&&d[i]<minT) minI=i,minT=d[i];
}
}
//if(minI>=0) cout<<"go to "<<minI<<" time_:"<<d[minI]<<endl;
s=minI;
if(s<0) break;
}
return d[r];
}
int main(){
int casen;
cin>>casen;
typedef pair<int,int> Pair;
for(int casei=1;casei<=casen;casei++){
memset(time_,-1,sizeof(time_));
memset(wait,0,sizeof(wait));
vector<vector<pair<int,int> > >tunes(1001,vector<Pair>());
int N;cin>>N;
//cout<<N<<" lines"<<endl;
int num=0;
for(int i=0;i<N;i++){
int ni,wi;
cin>>ni>>wi;
wait[i]=wi;
assert(wait[i]>=0);
//cout<<ni<<" stations, wait: "<<wait[i]<<endl;
start[i]=num,num+=ni;
for(int j=0;j<ni-1;j++){
int l=start[i]+j,r=start[i]+j+1;
int a;cin>>a;
//cout<<l<<": "<<r<<"="<<a<<endl;
time_[l][r]=time_[r][l]=a;
}
}
int M;cin>>M;
//cout<<M<<" tunes"<<endl;
for(int i=0;i<M;i++){
int l1,s1,l2,s2,ti;
cin>>l1>>s1>>l2>>s2>>ti;
l1--,l2--,s1--,s2--;
assert(l1!=l2);
int l=start[l1]+s1,r=start[l2]+s2;
tunes[l].push_back(Pair(r,l2));
tunes[r].push_back(Pair(l,l1));
time_[l][r]=ti+wait[l2];
time_[r][l]=ti+wait[l1];
//cout<<l<<": "<<r<<"="<<time_[l][r]<<" wait[r]:"<<wait[l2]<<endl;
}
int q;cin>>q;
printf("Case #%d:\n",casei);
//cout<<q<<" queies"<<endl;
floyd(num);
for(int i=0;i<q;i++){
int l1,s1,l2,s2;
cin>>l1>>s1>>l2>>s2;
l1--,l2--,s1--,s2--;
int l=start[l1]+s1,r=start[l2]+s2;
assert(l!=r);
int s=d[l][r]>=0?d[l][r]+wait[l1]:-1;
for(int j=0;j<tunes[l].size();j++){
Pair & k=tunes[l][j];
int t=wait[k.second];
cout<<"tune from "<<l<<" to "<<k.first<<endl;
if(s<0||d[k.first][r]>=0&&s>t+d[k.first][r])
s=t+d[k.first][r];
}
//cout<<l<<":"<<r<<endl;
printf("%d\n",s);//+wait[l1]);
}
}
return 0;
}
上题答案存在多处问题：
1 没考虑可以沿着隧道从起点走到终点
2 没考虑在隧道里走a->b->c.
上题目，先沿相同线枚举最短路径。然后中转站之间按照wait时间。然后floyd。
#include<cstdio>
#include <assert.h>
#include<cstring>
#include<cmath>
#include <vector>
#include <iostream>
using namespace std;
int time_[1001][1001];
// int wait[1001];
int start[1001];
int d[1001][1001];
int lst[1001][1001];
void floyd(int n){
memset( d, -1, sizeof( d ));
memset( lst, -1, sizeof( lst ));
for( int i = 0; i < n; i++ )
for( int j = 0; j < n; j++ )
d[i][j] = time_[i][j],
lst[i][j] = j;
for( int i = 0; i < n; i++ )
for( int j = 0; j < n; j++ )
for( int k = 0 ; k < n ; k++ )
if( d[j][i] >= 0 && d[i][k] >= 0 )
if( d[j][k] <0 || d[j][k] > d[j][i] + d[i][k] )
d[j][k] = d[j][i] + d[i][k],
lst[j][k] = lst[j][i];
}
int dkl(int l,int r,int n){
bool visit[1001];
int d[1001], lst[1001];
memset( d, -1, sizeof( d ) );
memset( lst, -1, sizeof( lst ) );
memset( visit, 0, sizeof( visit ) );
d[l] = 0;
int s = l;
while(s != r){
visit[s] = true;
int minT = 99999999, minI = -1;
for( int i = 0; i < n; i ++ ){
if( !visit[i] ){
if( time_[s][i] >= 0 && ( d[i]<0 || d[s] + time_[s][i] < d[i] ) ){
d[i] = d[s] + time_[s][i];
lst[i] = s;
}
if( d[i] >= 0 && d[i] < minT ) minI = i, minT = d[i];
}
}
s = minI;
if( s < 0 ) break;
}
/*
int i = s;
printf( "path from %d to %d : \n", l, r);
while( true ){
cout<<i<<" ";
i = lst[i];
if( i < 0 ) break;
}
cout<<endl;
cout<<"d[r]:"<<d[r]<<endl;
*/
return d[r];
}
int main(){
int casen; cin >> casen;
typedef pair < int, int > Pair;
for( int casei = 1; casei <= casen; casei++ ){
memset( time_, -1, sizeof(time_) );
int N; cin >> N;
int num=0;
// input line stations
for( int i = 0 ; i < N ; i++ ){
int ni, wi;
cin >> ni >> wi;
// wait[i] = wi;
start[i] = num, num += ni;
for( int j = 0 ; j < ni - 1 ; j++ ){
int l=start[i]+j,r=start[i]+j+1;
int a; cin >> a;
time_[l][r] = time_[r][l] = a;
}
for( int j = 0 ; j < ni ; j++ ){ // @error: j < ni - 1, copied from above, is wrong logic for (j,k) enumeration
for( int k = j + 2 ; k < ni ; k++ ){// @error: k < ni - 1, copied from above, is wrong logic for (j,k) enumeration
int l = start[i] + j, r = start[i] + k;
time_[r][l] = time_[l][r] = time_[l][r-1] + time_[r-1][r];
}
}
for( int j = 0 ; j < ni ; j++ ){ // @error: j < ni - 1, copied from above, is wrong logic for (j,k) enumeration
for( int k = j + 1 ; k < ni ; k++ ){// @error: k < ni - 1, copied from above, is wrong logic for (j,k) enumeration
// @error: int k = j + 2, copied from above, is wrong logic for (j,k)
int l = start[i] + j, r = start[i] + k;
time_[r][l] = time_[l][r] = time_[l][r] + wi;
}
}
}
// input tune stations
int M;cin >> M;
for( int i = 0 ;i < M ; i ++ ){
int l1, s1, l2, s2, ti;
cin >> l1 >> s1 >> l2 >> s2 >> ti;
l1--,l2--,s1--,s2--; assert(l1!=l2);
int l = start[l1] + s1, r = start[l2] + s2;
assert( l1 != l2 );
time_[l][r] = time_[r][l] = time_[l][r] < 0 ? ti : min( time_[l][r], ti );
}
// input queries
int q; cin >> q;
//// for debug floyd( num );
printf( "Case #%d:\n", casei );
for( int i = 0 ; i < q ; i ++ ){
int l1,s1,l2,s2;
cin >>l1>>s1>>l2>>s2;
l1--, l2--, s1--, s2--;
int l = start[l1] + s1, r = start[l2] + s2;
int dst = dkl( l, r, num );
printf( "%d\n", dst );
////assert( dst == d[l][r] );
}
}
return 0;
}
第三题：dp。
a[n]=MAXN or min(a[k]+a[n/k]+1，len(n) if n can be input by keyboard).
答案是a[n]+1 or impossible注意如果一个数可以直接用键盘打，就不要用乘法间接算了。
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int a[1000005],b[10];
int check( int x )
{
int ans = 0;
while ( x != 0 )
{
if ( b[x%10] == 0 ) return 0;
ans++;
x = x / 10;
}
return ans;
}
int main()
{
int cas;
int c = 0;
scanf( "%d", &cas );
while ( cas-- )
{
for ( int i = 0; i <= 1000000; i++ )
a[i] = 0;
int n;
for ( int i = 0; i < 10; i++ ) b[i] = 0;
for ( int i = 0; i < 10; i++ )
{
scanf( "%d", &n );
if ( n == 1 ) a[i] = 1;
if ( n == 1 ) b[i] = 1;
}
scanf( "%d", &n );
for ( int i = 2; i <= n; i++ )
{
if ( a[i] != 0 ) continue;
a[i] = check(i);
int p = int(sqrt(i));
for ( int j = 2; j <= p; j++ )
if ( (i % j) == 0 && a[j] != 0 && a[i/j] != 0 )
{
if ( a[i] == 0 ) a[i] = a[j] + a[i/j] + 1;
else if ( a[i] > a[j] + a[i/j] + 1 ) a[i] = a[j] + a[i/j] + 1;
}
}
c++;
printf("Case #%d: ",c);
if ( a[n] != 0 ) printf("%d\n",a[n]+1);
else printf("Impossible\n");
}
return 0;
}
第四题 为什么a不了？
#include<iostream>
#include<bitset>
using namespace std;
int matrix[104][104];
int sum[104];
int W;
int H;
int N;
int t;
int r;
int tet_x;
int flag = 1;
int tet[8][4];
int line[8][4];
void init(){
tet[1][0] = 0x8c4;
line[1][0] = 3;
tet[1][1] = 0x6c;
line[1][1] = 2;
tet[1][2] = 0x8c4;
line[1][2] = 3;
tet[1][3] = 0x6c;
line[1][3] = 2;
tet[2][0] = 0x4c8;
line[2][0] = 3;
tet[2][1] = 0xc6;
line[2][1] = 2;
tet[2][2] = 0x4c8;
line[2][2] = 3;
tet[2][3] = 0xc6;
line[2][3] = 2;
tet[3][0] = 0x88c;
line[3][0] = 3;
tet[3][1] = 0x2e;
line[3][1] = 2;
tet[3][2] = 0xc44;
line[3][2] = 3;
tet[3][3] = 0xe8;
line[3][3] = 2;
tet[4][0] = 0x44c;
line[4][0] = 3;
tet[4][1] = 0xe2;
line[4][1] = 2;
tet[4][2] = 0xc88;
line[4][2] = 3;
tet[4][3] = 0x8e;
line[4][3] = 2;
tet[5][0] = 0xcc;
line[5][0] = 2;
tet[5][1] = 0xcc;
line[5][1] = 2;
tet[5][2] = 0xcc;
line[5][2] = 2;
tet[5][3] = 0xcc;
line[5][3] = 2;
tet[6][0] = 0x8888;
line[6][0] = 4;
tet[6][1] = 0xf;
line[6][1] = 1;
tet[6][2] = 0x8888;
line[6][2] = 4;
tet[6][3] = 0xf;
line[6][3] = 1;
tet[7][0] = 0x4e;
tet[7][1] = 0x4c4;
tet[7][2] = 0xe4;
tet[7][3] = 0x8c8;
int a=42486;
bitset<32> bs(a);
/*
for(int i = 1; i<8;i++){
for(int j = 0;j<4;j++){
cout<<"here is the tet "<<i<<" "<<j<<endl;
bitset<32> bs(tet[i][j]);
cout<<bs<<endl;
for(int i = 3;i>=0;i--){
cout<<bs[4*i+3]<<bs[4*i+2]<<bs[4*i+1]<<bs[4*i]<<endl;
}
}
}
*/
}
void printMatrix(){
if(flag == 1){
for(int i = H-1;i>=0;i--){
for(int j=0; j<W;j++){
if(matrix[i][j] == 1){
cout<<"x";
}else{
cout<<".";
}
}
// cout<<" sum is: "<<sum[i]<<endl;
cout<<endl;
}
}else{
cout<<"Game Over!"<<endl;
}
}
void moveMatrix(int y){
int index = 0;
for(int i = y; i<H; i++){
if(sum[i] == W){
index ++;
for(int j = i+1;j<H;j++){
if(sum[j] == W)
index ++;
}
}
if(index != 0){
for(int j = 0;j<W;j++){
matrix[i][j] = matrix[i+index][j];
}
sum[i] = sum[i+index];
}
}
}
void putTet(int y,int x,int right){
if(right == 0){
return;
}
for(int i=0;i<4;i++){
int a = 1<<i;
int notZero = (right & a)>>i;
// cout<<"here a is: "<<a<<"here notZero is: "<<notZero<<endl;
if(notZero == 1){
matrix[y][x+3-i] = notZero;
sum[y]++;
// if(sum[y]>W){
// printMatrix();
// system("pause");
// }
// cout<<"here y is: "<<y<<" "<<"here sum is:"<<sum[y]<<endl;
// for(int i = 0;i<H;i++){
//
// for(int j =0; j<W;j++){
// cout<<matrix[i][j]<<" ";
// }
// cout<<endl;
// }
}
}
}
int start(int type , int rotation, int my_x){
int my_tet = tet[type][rotation];
int isStop = 0;
int loc_flag = 1;
int y;
int x;
for(y = H-1, x= my_x; y>=0; y--){
isStop = 0;
for(int j = 0; j<4; j++){
int temp = (matrix[y+j][x]<<3)|(matrix[y+j][x+1]<<2)|(matrix[y+j][x+2]<<1)|(matrix[y+j][x+3]);//y 能不能放
isStop = isStop|(temp<<j);
}
isStop = my_tet&isStop;
if(isStop !=0 ){
break;
}
}
if(y == H-1){// H
loc_flag = 0;
}else{
y++;
int right[4];
right[0] = (my_tet&0x000f);
right[1] = (my_tet&0x00f0)>>4;
right[2] = (my_tet&0x0f00)>>8;
right[3] = (my_tet&0xf000)>>12;
for(int i=0;i<4;i++){
/* if(y+i>=H && right[i]!=0){
loc_flag = 0;
break;
}
*/
// cout<<"here right is: "<<right[i]<<endl;
putTet(y+i,my_x,right[i]);
}
moveMatrix(y);
for(int i=H;i<H+4;i++){
if(sum[i]!=0) {
loc_flag=0;
break;
}
}
}
return loc_flag;
}
int main(){
int T;
init();
cin >>T;
for(int i=0;i<T;i++){
cin >> W >>H >>N;
flag = 1;
for(int i = 0;i<H+4;i++){
for(int j = 0;j<W+4;j++){
matrix [i][j] = 0;
}
sum [i] = 0;
}
for(int i=0;i<N;i++){
cin>>t>>r>>tet_x;
if(flag == 1){
flag = start(t,r,tet_x);
}else{
continue;
}
// printMatrix();
}
cout<<"Case #"<<i+1<<":"<<endl;
printMatrix();
}
// cout<<endl<<"end"<<endl;
// cout<<"dadad: "<<sum[0]<<endl;
// system("pause");
return 1;
}