Title Link: BZOJ-1006
Problem analysis
This problem is a bare model of the minimum number of staining figures for a chord graph.
The minimum coloring method of a string graph is to find out the perfect elimination sequence (MCS algorithm) of the chord graph, and then, in accordance with the perfect elimination sequence, backward forward from backward to the smallest color that each Dianran can dye.
The number of colors found is the smallest stain, but also the largest group.
Code
#include <iostream> #include <cstdlib> #include <cstdio> #include <cmath> #include <cstring > #include <algorithm> #include <queue>using namespace std;const int MAXN = 10000 + 5, MAXM = 1000000 + 5;int N, M, Ans;int V[MAXN], A[MAXN], COL[MAXN], used[maxn];bool visit[maxn];struct edge{int V; Edge *next;} E[MAXM * 2], *p = E, *point[maxn];inline void Addedge (int x, int y) {++p; P-v = y; P-Next = point[x]; POINT[X] = P;} struct Es{int p, q;es () {}es (int a, int b) {p = A; q = b;}}; struct Cmp{bool operator () (ES E1, es E2) {return e1.q < e2.q;}}; Priority_queue<es, Vector<es>, cmp> Q;//mcs the perfect elimination sequence void MCS () {for (int i = 1; I <= n; ++i) {v[i] = 0; Visit[i] = false;} while (! Q.empty ()) Q.pop (); Q.push (ES (1, 0)); int x, y;for (int i = n; i >= 1; i.) {while (true) {x = Q.top (). P; Q.pop (); if (! VISIT[X]) break; A[i] = x; VISIT[X] = true;for (Edge *j = point[x]; j; j = J-Next) {y = J-v;if (Visit[y]) continue;++v[y]; Q.push (ES (y, v[y]));}}} void Min_paint () {Ans = 1;int x;memset (col, 0, sizeof (COL)), memset (used, 0, sizeof (used)), for (int i = n; i >= 1; i.) {for (Edge *j = Point[a[i]]; j; j = J-Next) Used[col[j-v]] = I;x = 1;while (used[x] = = i) {++x;if (x > Ans) Ans = x;} Col[a[i]] = x;}} int main () {scanf ("%d%d", &n, &m); int A, b;for (int i = 1; I <= m; ++i) {scanf ("%d%d", &a, &b); Addedge (A, b); Addedge (b, a);} MCS (); Min_paint ();p rintf ("%d\n", Ans); return 0;}
[Bzoj 1006] [HNOI2008] The magical kingdom "minimum coloring of the chord graph"