FZU - 2254 英语考试

这个问题可以转化为：找一个连通图的最小生成树。

因为有自环，所以用prim算法会更简单一点。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>

using namespace std;
typedef long long ll;

const int MAXN = 1000 + 2;
const int MAXL = 10 + 2;

int n, m, k;
ll w;
typedef struct Line{
	int v;	int weight;
	Line(int a, int c){
		v = a, weight = c;
	}
	bool operator < (const Line &rhs) const{
		return weight > rhs.weight;
	}
}Line;

int vis[MAXN];
char str[MAXN][MAXL];
vector<Line> g[MAXN];

int str_cnt(char *a, char *b, int len){
	ll cnt = 0;
	for(int i = 0; i < len; i++)	if(a[i] != b[i])	cnt++;
	return cnt;
}

bool check(int u, int v){
	if(!vis[u]){
		vis[u] = 1;	return true;
	}
	if(!vis[v]){
		vis[v] = 1;	return true;
	}
	return false;
}

ll MP(){
	ll res = m;
	memset(vis, 0, sizeof(vis));	priority_queue<Line> q;
	vis[1] = 1;
	for(int i = 0; i < g[1].size(); i++){
		int v = g[1][i].v;	if(vis[v])	continue;
		q.push(g[1][i]), q.push(Line(v, m));
	}
	while(!q.empty()){
		Line cur = q.top();	q.pop();
		int u = cur.v;	ll weight = cur.weight;
		if(vis[u])	continue;
		res += weight;	vis[u] = 1;
		for(int i = 0; i < g[u].size(); i++){
			int v = g[u][i].v;
			if(vis[v])	continue;
			q.push(g[u][i]), q.push(Line(v, m));
		}
	}
	return res;
}

int main(void)
{
	while(~scanf("%d%d%lld", &n, &m, &w)){
		for(int i = 1; i <= n; i++)	g[i].clear();
		for(int i = 1; i <= n; i++)	scanf("%s", str[i]);
		for(int i = 1; i <= n; i++){
			for(int j = i + 1; j <= n; j++){
				g[i].push_back(Line(j, str_cnt(str[i], str[j], m) * w));
				g[j].push_back(Line(i, str_cnt(str[i], str[j], m) * w));
			}
		}
		printf("%lld\n", MP());
	}return 0;
}