poj 1129 Channel Allocation (四色定理)

http://poj.org/problem?id=1129

四色定理

#include "stdio.h" #include "iostream" #include "cstring" using namespace std; bool g[27][27]; int color[27]; bool dfs(int nColor, int now, int n) //nColor:所用颜色总数；now:当前待填颜色的物品的编号 { if(now > n) return true; for(int i = 1; i <= nColor; i++) { bool flag = true; color[now] = i; //尝试颜色i for(int j = 1; j <= n; j++) { if(g[now][j] && color[j] == i) //相邻且着色一样时 { flag = false; break; } } if(flag && dfs(nColor, now + 1, n)) return true; } return false; } int main() { int n, i, j; char str[40]; while(scanf("%d", &n) && n) { memset(g, false, sizeof(g)); memset(color, -1, sizeof(color)); for(i = 1; i <= n; i++) { scanf("%s", str); j = 2; while(str[j]) { g[i][str[j] - 'A' + 1] = true; j++; } } //由四色定理 for(int nColor = 1; nColor <= 4; nColor++) { if(dfs(nColor, 0, n)) { if(nColor == 1) cout << nColor << " channel needed."<< endl; else cout << nColor << " channels needed."<< endl; break; } } } return 0; }