UVa 10004 Bicoloring

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本文介绍了一种用于解决UVa10004Bicoloring问题的图染色算法实现。通过广度优先搜索（BFS）策略判断图是否可以进行二染色，并提供了完整的C++代码示例。该算法适用于解决图论中的二部图判断问题。

/* File: 10004.cpp Author: ACboy Date: 2010-4-26 Result: 1A Descripition: UVa 10004 Bicoloring */ #include <iostream> using namespace std; int edges[210][210]; /* adjacency information */ int degree[210]; /* outdegree of each vertices */ int nvertices; /* number of vertices of graph */ int nedges; /* number of edges of graph */ int colors[210]; /* the colors of the each vertices*/ int vis[210]; /* flag the vertices if visited */ void bfs() { int queue[210]; int front; int rear; front = 0; rear = 0; memset(colors, 0, sizeof(colors)); memset(vis, 0, sizeof(vis)); vis[0] = 1; queue[front++] = 0; int c = 0; // 用来标记被成功染色的顶点的个数 while (rear < front) { int v = queue[rear++]; int i; int flag1 = 0; /* 用来标记顶点v的邻接点中是否有染色为1的点 */ int flag2 = 0; /* 用来标记顶点v的邻接点中是否有染色为2的点 */ for (i = 0; i < degree[v]; i++) { if (colors[edges[v][i]] == 1) { flag1 = 1; } if (colors[edges[v][i]] == 2) { flag2 = 1; } } if (flag2 == 0 && flag1 == 1) { /* 顶点v邻接点中只有染色为1的点*/ colors[v] = 2; /* 将顶点v染色为2*/ c++; /* 成功染色的顶点个数加一 */ } if (flag1 == 0 && flag2 == 1) { colors[v] = 1; c++; } if (flag1 == 0 && flag2 == 0) { colors[v] = 1; c++; } for (i = 0; i < degree[v]; i++) { if (vis[edges[v][i]] == 0) { queue[front++] = edges[v][i]; vis[edges[v][i]] = 1; } } } if (c == nvertices) { cout << "BICOLORABLE." << endl; } else { cout << "NOT BICOLORABLE." << endl; } } int main() { #ifndef ONLINE_JUDGE freopen("10004.txt", "r", stdin); #endif while (cin >> nvertices) { if (nvertices == 0) break; else { cin >> nedges; int i; int x, y; memset(degree, 0, sizeof(degree)); for (i = 0; i < nedges; i++) { cin >> x >> y; edges[x][degree[x]] = y; degree[x]++; edges[y][degree[y]] = x; degree[y]++; } bfs(); } } return 0; }