CF662B Graph Coloring 題解 搜尋
阿新 • • 發佈:2020-10-29
題目連結:https://www.luogu.com.cn/problem/CF662B
題目大意:
有 \(n\) 個點 \(m\) 條邊,邊有顏色 —— 紅色或藍色。你可以對點進行操作,每操作一個點,與這個點相鄰的所有邊會變化顏色(紅色變藍色,藍色變紅色)。問:最少需要操作幾次能夠讓邊的顏色都一樣(並輸出最少需要操作的所有點的編號)。
解題思路:
可能存在多個連通塊。首先我需要列舉將要變成的顏色,然後對於每一個連通塊中第一個碰到的點,我列舉這個點要不要進行操作。
因為一條邊連線兩個點,這個點有沒有操作直接影響了和它相鄰的邊要不要操作。同時一個點最多操作依次(操作兩次就跟沒操作一樣了)。
所以,總的時間複雜度為 \(O(4 \cdot n)\)
示例程式碼:
#include <bits/stdc++.h> using namespace std; const int maxn = 100010; struct Edge { int u, v, c, cc, nxt; // cc表示原始的顏色 Edge() {} Edge(int _u, int _v, int _c, int _nxt) { u = _u; v = _v; c = cc = _c; nxt = _nxt; } } edge[maxn<<1]; int head[maxn], n, m, ecnt, tmp, sum[2]; vector<int> res[2], tmp_vec, vecs[2]; void init() { ecnt = 0; memset(head, -1, sizeof(int)*(n+1)); } void addedge(int u, int v, int c) { edge[ecnt] = Edge(u, v, c, head[u]); head[u] = ecnt ++; edge[ecnt] = Edge(v, u, c, head[v]); head[v] = ecnt ++; } bool vis[maxn]; bool dfs(int u, bool paint, int c) { vis[u] = true; if (paint) { tmp ++; tmp_vec.push_back(u); for (int i = head[u]; i != -1; i = edge[i].nxt) { edge[i].c ^= 1; edge[i^1].c ^= 1; } } for (int i = head[u]; i != -1; i = edge[i].nxt) { int v = edge[i].v; if (edge[i].c != c) { if (vis[v]) return false; if (!dfs(v, true, c)) return false; } else if (!vis[v] && !dfs(v, false, c)) return false; } return true; } void recover(int u) { vis[u] = false; for (int i = head[u]; i != -1; i = edge[i].nxt) { edge[i].c = edge[i^1].c = edge[i].cc; int v = edge[i].v; if (vis[v]) recover(v); } } void handle(int u, int c, int id) { tmp = 0; int tmp1 = -1; for (int i = 0; i < 2; i ++) vecs[i].clear(); tmp_vec.clear(); if (dfs(u, true, c)) { tmp1 = tmp; for (int i = 0; i < tmp1; i ++) vecs[0].push_back(tmp_vec[i]); } recover(u); tmp = 0; int tmp2 = -1; tmp_vec.clear(); if (dfs(u, false, c)) { tmp2 = tmp; for (int i = 0; i < tmp2; i ++) vecs[1].push_back(tmp_vec[i]); } if (tmp1 == -1 && tmp2 == -1 ) sum[id] = -1; else if (tmp2 == -1 || tmp1 != -1 && tmp1 < tmp2) { sum[id] += tmp1; for (int i = 0; i < tmp1; i ++) res[id].push_back(vecs[0][i]); } else { sum[id] += tmp2; for (int i = 0; i < tmp2; i ++) res[id].push_back(vecs[1][i]); } } int main() { scanf("%d%d", &n, &m); init(); while (m --) { int u, v, c; char cc[2]; scanf("%d%d%s", &u, &v, cc); c = (cc[0] == 'R'); addedge(u, v, c); } for (int i = 1; i <= n && sum[0] != -1; i ++) { if (!vis[i]) handle(i, 0, 0); } for (int i = 1; i <= n; i ++) vis[i] = false; for (int i = 0; i < ecnt; i ++) edge[i].c = edge[i].cc; for (int i = 1; i <= n && sum[1] != -1; i ++) { if (!vis[i]) handle(i, 1, 1); } if (sum[0] == -1 && sum[1] == -1) puts("-1"); else if (sum[1] == -1 || sum[0] != -1 && sum[0] < sum[1]) { printf("%d\n", sum[0]); for (int i = 0; i < sum[0]; i ++) printf("%d ", res[0][i]); } else { printf("%d\n", sum[1]); for (int i = 0; i < sum[1]; i ++) printf("%d ", res[1][i]); } return 0; }