一共只有三種顏色,假設前兩種顏色反向,對於第三種顏色,它與前兩種齒輪中必有一種同向。
枚舉兩種齒輪的顏色,假設它們旋轉方向矛盾(同向),另一種只需要與它們反向就一定不矛盾了,所以此處可以不考慮。
對於這兩種需要齧合的顏色齒輪連邊,Hungary求二分圖最大匹配,最大匹配數就是矛盾需刪除掉的齒輪數(注:刪掉的不一定是同色齒輪)。
Code:
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<vector>
#include<algorithm>
using namespace std;
const int Max = 50;
struct node{
int v, nxt;
}edge[Max * Max + 5];
int N, cnt, Ans = 0x3f3f3f3f;
int fir[Max + 5], match[Max + 5];
bool vis[Max + 5];
char Col[Max + 5];
char Ori[3] = {'R', 'G', 'B'};
char map[Max + 5][Max + 5];
void addedge(int a, int b){
edge[++ cnt].v = b, edge[cnt].nxt = fir[a], fir[a] = cnt;
}
bool Dfs(int u){
for(int i = fir[u]; i; i = edge[i].nxt) if(! vis[edge[i].v]){
vis[edge[i].v] = 1;
if(! match[edge[i].v] || Dfs(match[edge[i].v])){
match[edge[i].v] = u;
return 1;
}
}
return 0;
}
void Hungary(){
int now = 0;
memset(match, 0, sizeof match );
for(int i = 1; i <= N; ++ i){
memset(vis, 0, sizeof vis );
now += Dfs(i);
}
Ans = min(Ans, now);
}
int main(){
//freopen("data1.in", "r", stdin);
freopen("gear.in", "r", stdin);
freopen("gear.out", "w", stdout);
scanf("%s", Col + 1);
N = strlen(Col + 1);
for(int i = 1; i <= N; ++ i)
scanf("%s", map[i] + 1);
for(int a = 0; a < 3; ++ a)
for(int b = a + 1; b < 3; ++ b){
memset(fir, 0, sizeof fir );
cnt = 0;
for(int i = 1; i <= N; ++ i) if(Col[i] == Ori[a]){
for(int j = 1; j <= N; ++ j) if(Col[j] == Ori[b])
if(map[i][j] == 'Y')
addedge(i, j);
}
Hungary();
}
printf("%d\n", Ans);
return 0;
}