[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;}

转载于:https://www.cnblogs.com/jffifa/archive/2012/03/21/2410295.html