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题意:要求A:BCD,A与B,C,D都不相同,求不同的值,典型的四色定理;
#include <iostream> #include <algorithm> #include <stdlib.h> #include <time.h> #include <cmath> #include <cstdio> #include <string> #include <cstring> #include <vector> #include <queue> #include <stack> #include <set> #define c_false ios_base::sync_with_stdio(false); cin.tie(0) #define INF 0x3f3f3f3f #define INFL 0x3f3f3f3f3f3f3f3f #define zero_(x,y) memset(x , y , sizeof(x)) #define zero(x) memset(x , 0 , sizeof(x)) #define MAX(x) memset(x , 0x3f ,sizeof(x)) #define swa(x,y) {LL s;s=x;x=y;y=s;} using namespace std ; #define N 105 const double PI = acos(-1.0); typedef long long LL ; int mapp[N][N], color[N]; int col,flag; int n; string s; bool ok(int i){ for(int j = 1; j <= 26; j++){ if(!mapp[i][j]) continue; if(color[i] == color[j]) return false; } return true; } void dfs(int num){ if(num > n) {flag = 1; return ;} for(int i = 1; i <= col; i ++){ color[num] = i; if(ok(num)) dfs(num+1); color[num]= 0; ///记得回溯时要清零啊!!! } } int main(){ //freopen("in.txt","r",stdin); //freopen("out.txt","w",stdout); while(cin>>n && n){ flag = 0; zero(mapp); for(int i = 0; i< n;i++){ cin>>s; int k = s.size(); int x,y; for(int j = 2;j< k;j++){ x = s[0] - ‘A‘ +1; y = s[j] - ‘A‘ +1; mapp[x][y] = mapp[y][x] = 1; } } for(col = 1; col <= 4; col++){ dfs(1); if(flag) break; } if(col == 1) printf("%d channel needed.\n", col); else printf("%d channels needed.\n", col); } return 0; }
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原文地址:http://www.cnblogs.com/yoyo-sincerely/p/5294285.html