[解题报告]Codeforces 105D Entertaining Geodetics

Abstract

Codeforces 105D

并查集（官方标的）

Body

Source

http://codeforces.com/problemset/problem/105/D

Description

题意很复杂，自己看……我一开始也没看懂，把那个gif动态图看个10遍以上就懂了。

http://212.193.37.254/codeforces/images/geo_slow.gif

Solution

比较裸的并查集。注意到染色操作实际上就是将原本颜色不同的格子合并到一起即可。计算螺旋形距离的话，实际上只跟格子相对中心格的坐标有关。我是直接找公式，CF上面的代码好像大部分是直接记录，各种step++什么的，看不懂。

但是仔细分析后我觉得，其实和并查集没有什么关系……令当前需要染色的格子集合为S。可以证明，实际上每次染色只会把别的格子并入S，而不会从S中删除或换掉格子（尽管S的颜色不断改变）。于是只用记录S的颜色和大小即可。

P.S. 写得这么简陋还叫解题报告实在是太不好意思了，以后标签改成[代码]好了。

Code

并查集（我写的并查集很长很难看，大家还是看CF上面的代码好了）

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <queue>
#include <map>
using namespace std;

const int S = 303*303*2;

typedef long long LL;
typedef pair<int, int> pr;
pr (*mkpr)(int, int)=make_pair;


int N, M;
LL ans = 0;
map<int, int> color;
int f[S], r[S], c[S], s[S];
int pan[303][303], sym[303][303];
int colorcnt = 0;
vector<pr> node[S];
queue<pr> q;
pr now;
int cnow, cchg;

inline int getcolor(int x)
{
    if (!color.count(x)) color[x] = colorcnt++;
    return color[x];
}

inline int dis(int i, int j)
{
    int n;
    if (j <= 0) n = max(abs(j), abs(i)-1);
    else n = max(abs(j), abs(i))-1;
    if (i==n+1) return 4*(n+1)*(n+1)+(n+1)-j;
    else if (j==n+1) return (4*n+2)*(n+1)+(n+1)+i;
    else if (i==-n-1) return (2*n+1)*(2*n+1)+n+j;
    else return 4*n*n+2*n+n-i;
}

bool cmp(const pr &u, const pr &v)
{return dis(u.first-now.first, u.second-now.second)<dis(v.first-now.first, v.second-now.second);}

int find(int x)
{
    if (f[x]==x) return x;
    return f[x] = find(f[x]);
}

int join(int u, int v)
{
    int ru = find(u), rv = find(v);
    if (ru == rv) return ru;
    if (r[ru]<r[rv])
    {
        s[rv] += s[ru];
        s[ru] = 0;
        return f[ru] = rv;
    }
    else if (r[rv]<r[ru])
    {
        s[ru] += s[rv];
        s[rv] = 0;
        return f[rv] = ru;
    }
    else
    {
        r[rv]++;
        s[rv] += s[ru];
        s[ru] = 0;
        return f[ru] = rv;
    }
}

void init()
{
    for (int i = 0; i < colorcnt; ++i)
    {
        f[i] = i;
        c[i] = i;
        r[i] = 0;
    }
}

int main()
{
    int i, j, k, x, y, root;
    scanf("%d%d", &N, &M);
    color[0] = colorcnt++;
    for (i = 0; i < N; ++i)
        for (j = 0; j < M; ++j)
        {
            scanf("%d", &pan[i][j]);
            pan[i][j] = getcolor(pan[i][j]);
            s[pan[i][j]]++;
        }
    for (i = 0; i < N; ++i)
        for (j = 0; j < M; ++j)
        {
            scanf("%d", &sym[i][j]);
            if (sym[i][j]!=-1) sym[i][j] = getcolor(sym[i][j]);
        }
    init();
    scanf("%d%d", &x, &y);
    x--; y--;
    for (i = 0; i < N; ++i)
        for (j = 0; j < M; ++j)
            if (i==x&&j==y) q.push(mkpr(i, j));
            else if (sym[i][j]!=-1) node[pan[i][j]].push_back(mkpr(i, j));
    while (!q.empty())
    {
        now = q.front(); q.pop();
        i = now.first, j = now.second;
        root = find(pan[i][j]);
        cnow = c[root]; cchg = sym[i][j];
        if (cnow==0 || cnow==cchg) continue;
        ans += s[root];
        vector<pr> elm = node[cnow];
        node[cnow].clear();
        sort(elm.begin(), elm.end(), cmp);
        for (k = 0; k < elm.size(); ++k)
            q.push(elm[k]);
        f[cchg] = cchg;
        c[join(root, cchg)] = cchg;
    }
    cout << ans << endl;
}

非并查集

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <queue>
#include <map>
using namespace std;

const int S = 303*303*2;

typedef long long LL;
typedef pair<int, int> pr;
pr (*mkpr)(int, int)=make_pair;


int N, M;
LL ans = 0;
map<int, int> color;
int s[S];
int pan[303][303], sym[303][303];
int colorcnt = 0;
vector<pr> node[S];
queue<pr> q;
pr now;
int snow, cnow, cchg;

inline int getcolor(int x)
{
    if (!color.count(x)) color[x] = colorcnt++;
    return color[x];
}

inline int dis(int i, int j)
{
    int n;
    if (j <= 0) n = max(abs(j), abs(i)-1);
    else n = max(abs(j), abs(i))-1;
    if (i==n+1) return 4*(n+1)*(n+1)+(n+1)-j;
    else if (j==n+1) return (4*n+2)*(n+1)+(n+1)+i;
    else if (i==-n-1) return (2*n+1)*(2*n+1)+n+j;
    else return 4*n*n+2*n+n-i;
}

bool cmp(const pr &u, const pr &v)
{return dis(u.first-now.first, u.second-now.second)<dis(v.first-now.first, v.second-now.second);}

int main()
{
    int i, j, k, x, y;
    scanf("%d%d", &N, &M);
    color[0] = colorcnt++;
    for (i = 0; i < N; ++i)
        for (j = 0; j < M; ++j)
        {
            scanf("%d", &pan[i][j]);
            pan[i][j] = getcolor(pan[i][j]);
            s[pan[i][j]]++;
        }
    for (i = 0; i < N; ++i)
        for (j = 0; j < M; ++j)
        {
            scanf("%d", &sym[i][j]);
            if (sym[i][j]!=-1) sym[i][j] = getcolor(sym[i][j]);
        }
    scanf("%d%d", &x, &y);
    x--; y--;
    for (i = 0; i < N; ++i)
        for (j = 0; j < M; ++j)
            if (i==x&&j==y) q.push(mkpr(i, j));
            else if (sym[i][j]!=-1) node[pan[i][j]].push_back(mkpr(i, j));
    cnow = pan[x][y]; snow = 0;
    while (!q.empty())
    {
        now = q.front(); q.pop();
        i = now.first, j = now.second;
        cchg = sym[i][j];
        if (cnow==0 || cnow==cchg) continue;
        snow += s[cnow];
        s[cnow] = 0;
        ans += snow;
        vector<pr> elm = node[cnow];
        node[cnow].clear();
        sort(elm.begin(), elm.end(), cmp);
        for (k = 0; k < elm.size(); ++k)
            q.push(elm[k]);
        cnow = cchg;
    }
    cout << ans << endl;
}