2014 ACM/ICPC Asia Regional Guangzhou Online, 2014 icpc

2014 ACM/ICPC Asia Regional Guangzhou Online, 2014 icpc

Question: 1002 A shopping upt Mayor's Performance Art

There is a sequence with n length. The initial staining is 2. There are two operations: P x, y, z, and the interval x --- y is z, and the other is Q x, y, the query interval x -- y has several colors and is output. Note that it will overwrite.

Analysis: it is the same as POJ 2777, but the color must be output. Therefore, when querying the line segment tree, save the path and print it. Code:

AC code:

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <vector>#include <map>using namespace std;const int N = 1110000;struct Node{    int l,r;    long long num;};Node tree[4*N];map<int,int> mp;//int vis[35];vector<int> v;void build(int l,int r,int o){    tree[o].l=l;    tree[o].r=r;    tree[o].num=2;    if(l==r)        return ;    int mid=(l+r)/2;    build(l,mid,o*2);    build(mid+1,r,o*2+1);}void update(int l,int r,int t,int o){    if(tree[o].l==l && tree[o].r==r)    {        tree[o].num=t;        return;    }    if(tree[o].num==t) return;    if(tree[o].num!=-1)    {        tree[2*o].num=tree[o].num;        tree[2*o+1].num=tree[o].num;        tree[o].num=-1;    }    int mid=(tree[o].l+tree[o].r)>>1;    if(r<=mid)        update(l,r,t,o+o);    else if(l>mid)        update(l,r,t,o+o+1);    else    {        update(l,mid,t,o*2);        update(mid+1,r,t,o*2+1);    }}void query(int l,int r,int o){    if(tree[o].num!=-1)    {        if(!mp[tree[o].num]){            v.push_back(tree[o].num);            mp[tree[o].num]=1;        }        return ;    }    int mid=(tree[o].l+tree[o].r)>>1;    if(r<=mid)        query(l,r,o+o);    else if(l>mid)        query(l,r,o+o+1);    else    {        query(l,mid,o*2);        query(mid+1,r,o*2+1);    }}int main(){    //freopen("Input.txt","r",stdin);    int n,m;    while(~scanf("%d%d",&n,&m))    {        if(n==0 && m==0)            break;        build(1,n,1);        while(m--)        {            getchar();            char ok;            int x,y,z;            scanf("%c",&ok);            if(ok=='P')            {                scanf("%d%d%d",&x,&y,&z);                update(x,y,z,1);            }            else            {                int ans=0;                v.clear();                mp.clear();                scanf("%d%d",&x,&y);                query(x,y,1);                sort(v.begin(),v.end());                for(int i=0;i<v.size();i++)                    printf("%d%c",v[i],i==(v.size()-1)?'\n':' ');            }        }    }    return 0;}

Question: 1003 Wang Xifeng's Little Plot

Question: to give a picture, you can only turn 90 degrees. How much can you go?

Analysis: enumerate each vertex as a corner point, then judge the value in eight directions, save it, and then save a maximum value,

AC code:

#include <cstdio>#include <cstring>#include <algorithm>#include <iostream>#include <vector>#include <map>using namespace std;const int N = 111;char mp[N][N];int dis[10];int n;int solve(int x,int y){    memset(dis,0,sizeof(dis));    for(int j=y;j<n;j++)    {        if(mp[x][j]=='.')            dis[0]++;        else            break;    }    for(int i=x,j=y;i<n&&j<n;i++,j++)    {        if(mp[i][j]=='.')            dis[1]++;        else            break;    }    for(int i=x;i<n;i++)    {        if(mp[i][y]=='.')            dis[2]++;        else            break;    }    for(int i=x,j=y;i<n && j>=0;i++,j--)    {        if(mp[i][j]=='.')            dis[3]++;        else            break;    }    for(int j=y;j>=0;j--)    {        if(mp[x][j]=='.')            dis[4]++;        else            break;    }    for(int i=x,j=y;i>=0 && j>=0;i--,j--)    {        if(mp[i][j]=='.')            dis[5]++;        else            break;    }    for(int i=x;i>=0;i--)    {        if(mp[i][y]=='.')            dis[6]++;        else            break;    }    for(int i=x,j=y;i>=0 && j<n;i--,j++)    {        if(mp[i][j]=='.')            dis[7]++;        else            break;    }    int ans=0;    for(int i=0;i<8;i++)    {        ans=max(ans,dis[i]+dis[(i+2)%7]);        ans=max(ans,dis[i]+dis[(i+4)%7]);    }    return (ans-1);}int main(){    //freopen("Input.txt","r",stdin);    while(~scanf("%d",&n) && n)    {        for(int i=0;i<n;i++)        {            getchar();            for(int j=0;j<n;j++)                scanf("%c",&mp[i][j]);        }        int ans=0;        for(int i=0;i<n;i++)        {            for(int j=0;j<n;j++){                //printf("%c",mp[i][j]);                if(mp[i][j]=='.')                    ans=max(ans,solve(i,j));            }        }        printf("%d\n",ans);    }    return 0;}

Question: 1004 Saving Tang Monk

Question: Let's give a picture, from K to T, there are keys on the road, a maximum of 9, then there is a strange, killing the strange cost 1, you have to get all the keys, calculate the shortest time.

Analysis: BFS + state Compression

I started to think about the state compression key. I found that killing monsters also needs State compression. In this way, the Array Memory is too large, and I found that the key has a condition that I had to get all the first few keys to get the next one, then we can simply save it with the array [9], and use the priority queue because the kill has a cost.

AC code:

#include <cstdio>#include<iostream>#include <queue>#include <cstring>using namespace std;const int N = 100;char mp[N][N];int vis[N][N][10][1<<5];struct Node{    int x,y,step;    int key ,e ;};int dx[6]= {0,0,1,-1};int dy[6]= {1,-1,0,0};int m,n,t;bool operator <(Node a, Node b){    return a.step > b.step;}int BFS(Node st,Node en){    memset(vis,0,sizeof(vis));    st.step = 0;    st.key=0,st.e=0;    vis[st.x][st.y][0][0]=1;    priority_queue<Node> q;    q.push(st);    Node tmp1,tmp2;    while(!q.empty())    {        tmp1 = q.top();        //printf("xxx%d\n",tmp1.step);        q.pop();        //printf("---%d %dxx\n",tmp1.key ,(1<<(t)-1));        if(tmp1.x==en.x && tmp1.y==en.y && tmp1.key== t)            return tmp1.step;        for(int i=0; i<4; i++)        {            tmp2.x=tmp1.x+dx[i];            tmp2.y=tmp1.y+dy[i];            tmp2.step = tmp1.step+1;            tmp2.e = tmp1.e ;            tmp2.key = tmp1.key ;            if(tmp2.x >=1 && tmp2.x <= n && tmp2.y >= 1 && tmp2.y <= n && mp[tmp2.x][tmp2.y] != '#')            {                if(mp[tmp2.x][tmp2.y]>='A' && mp[tmp2.x][tmp2.y] <='F')                {                    //printf("No\n");                    int tu = mp[tmp2.x][tmp2.y]-'A' ;                    if(tmp2.e &(1<<tu))//                    {                        if(!vis[tmp2.x][                           tmp2.y][tmp2.key][tmp2.e])                        {                            vis[tmp2.x][tmp2.y][tmp2.key][tmp2.e] = true ;                            q.push(tmp2) ;                        }                    }                    else                    {                        tmp2.step++ ;                        tmp2.e |= (1<<tu) ;                        q.push(tmp2) ;                    }                }                else if(mp[tmp2.x][tmp2.y]>='1' && mp[tmp2.x][tmp2.y]<='9')                {                    //printf("YES\n");                    int tu = mp[tmp2.x][tmp2.y]-'0' ;                    if(tu == tmp2.key + 1)                    {                        tmp2.key = tu ;                        q.push(tmp2) ;                    }                    else                    {                        if(!vis[tmp2.x][tmp2.y][tmp2.key][tmp2.e])                        {                            vis[tmp2.x][tmp2.y][tmp2.key][tmp2.e] = true ;                            q.push(tmp2) ;                        }                    }                }                else if(!vis[tmp2.x][tmp2.y][tmp2.key][tmp2.e])                {                    vis[tmp2.x][tmp2.y][tmp2.key][tmp2.e] = true ;                    q.push(tmp2) ;                }            }        }    }    return -1;}int main(){    //freopen("Input.txt","r",stdin);    while(~scanf("%d%d",&n,&t))    {        char c='A';        if(n==0 && t==0)            break;        Node st,en;        m=n;        for(int i=1; i<=m; i++)        {            getchar();            for(int j=1; j<=n; j++)            {                scanf("%c",&mp[i][j]);                if(mp[i][j]=='S')                {                    mp[i][j]=c++;                }                if(mp[i][j]=='K')                    st.x=i,st.y=j;                if(mp[i][j]=='T')                    en.x=i,en.y=j;            }        }        int ans=BFS(st,en);//        for(int i=1;i<=n;i++)//        {//            for(int j=1;j<=n;j++)//            {//                printf("%d",vis[i][j][1]);//            }//            printf("\n");//        }        if(ans==-1)            printf("impossible\n");        else            printf("%d\n",ans);    }    return 0;}

2012 how many teams are ACM/ICPC Asia Regional Chengdu Online?

130 teams
 
What is ACM/ICPC Asia Regional?

International. The finals are world-class. Asia belongs to the intercontinental and international level!