USACO 5.1 Starry Night

这个题我直接模拟的，具体做法是：

找到每种图案所在的最小矩形块，把占得长和宽记录一下。然后以最小矩形块的左下角为原点，将团中的坐标重构并排序存储。然后不同图案直接对比，然后标号就行了。注意对比的时候需要考虑4种旋转，还有反转。总共8种情况。。。又写了160来行，160行已然无压力了。。。样例过了一般就过了吧。。。贴下挫代码：

/*
ID: zlqest11
LANG: C++
TASK: starry
*/
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;

const short N = 105;

int m, n, ct;
char str[N][N];
int lab[N][N];
int totind, ind[505];
struct node{
    int x, y;
    node(){}
    node(int a, int b){
        x = a, y = b;
    }
};
struct cluster{
    int h, w;
    vector <node> loc;
}pic[505];

bool cmp(const node &a, const node &b){
    if(a.x == b.x)
        return  a.y < b.y;
    else
        return  a.x < b.x;
}

void floodfill(int x, int y, int ct){
    queue <node> q;
    node u;
    int tx, ty;
    lab[x][y] = ct;
    q.push(node(x,y));
    while(!q.empty()){
        u = q.front();
        pic[ct].loc.push_back(node(u.x, u.y));
        q.pop();
        for(int px = -1; px <= 1; px++)
            for(int py = -1; py <= 1; py++){
                tx = u.x + px;
                ty = u.y + py;
                if(tx<0 || tx>=n || ty<0 || ty>=m)  continue;
                if(lab[tx][ty]==-1 && str[tx][ty]=='1'){
                    lab[tx][ty] = ct;
                    q.push(node(tx, ty));
                }
            }
    }
    int len = pic[ct].loc.size();
    int lx, ly, rx, ry;
    sort(pic[ct].loc.begin(), pic[ct].loc.end(), cmp);
    lx = ly = 150;
    rx = ry = -1;
    for(int i = 0; i < len; i++){
        lx = min(lx, pic[ct].loc[i].x);
        ly = min(ly, pic[ct].loc[i].y);
        rx = max(rx, pic[ct].loc[i].x);
        ry = max(ry, pic[ct].loc[i].y);
    }
    for(int i = 0; i < len; i++){
        pic[ct].loc[i].x -= lx-1;
        pic[ct].loc[i].y -= ly-1;
    }
    pic[ct].h = rx-lx+1;
    pic[ct].w = ry-ly+1;
}

void roll(int pos){
    node tmp;
    int len = pic[pos].loc.size();
    for(int i = 0; i < len; i++){
        tmp.y = pic[pos].h-pic[pos].loc[i].x+1;
        tmp.x = pic[pos].loc[i].y;
        pic[pos].loc[i] = tmp;
    }
    swap(pic[pos].h, pic[pos].w);
    sort(pic[pos].loc.begin(), pic[pos].loc.end(), cmp);
}

void flip(int pos){
    int len = pic[pos].loc.size();
    for(int i = 0; i < len; i++)
        pic[pos].loc[i].x = pic[pos].h-pic[pos].loc[i].x+1;
    sort(pic[pos].loc.begin(), pic[pos].loc.end(), cmp);
}

bool isSame(int a, int b){
    int i, len = pic[a].loc.size();
    for(i = 0; i < 4; i++){
        roll(b);
        bool flag = true;
        if(pic[b].h!=pic[a].h || pic[b].w!=pic[a].w)
            flag = false;
        if(pic[b].loc.size() != pic[a].loc.size())
            flag = false;
        for(int j = 0; j < len; j++){
            if(pic[a].loc[j].x!=pic[b].loc[j].x || pic[a].loc[j].y!=pic[b].loc[j].y){
                flag = false;
                break;
            }
        }
        if(flag)  return  true;
    }
    return  false;
}

int main()
{
    freopen("starry.in", "r", stdin);
    freopen("starry.out", "w", stdout);
    scanf("%d%d", &m, &n);
    for(int i = 0; i < n; i++)
        scanf(" %s", str[i]);
    memset(lab, -1, sizeof(lab));
    for(int i = ct = 0; i < n; i++)
        for(int j = 0; j < m; j++)
            if(lab[i][j]==-1 && str[i][j]=='1'){
                floodfill(i, j, ct);
                ++ct;
            }
    memset(ind, -1, sizeof(ind));
    totind = 0;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++){
            if(ind[lab[i][j]]==-1 && str[i][j]=='1'){
                ind[lab[i][j]] = ++totind;
                for(int k = 0; k < ct; k++){
                    if(isSame(lab[i][j], k)){
                        ind[k] = totind;
                    }
                    if(ind[k] != totind){
                        flip(k);
                        if(isSame(lab[i][j], k)){
                            ind[k] = totind;
                        }
                    }
                }
            }
        }
    for(int i = 0; i < n; i++)
        for(int j = 0; j < m; j++){
            if(lab[i][j]!=-1 && str[i][j]=='1'){
                str[i][j] = 'a'+ind[lab[i][j]]-1;
            }
        }
    for(int i = 0; i < n; i++)
        printf("%s\n", str[i]);
    return 0;
}