B

Think:
1知识点：并查集
2题意思考：将每一个单词视为一个结点，两点之间的权值即为联想背这两个单词的精力，这样背过n个单词即可认为将n个结点生成一棵树
3反思：初始化注意结点范围

vjudge题目链接

以下为Wrong Answer代码——初始化范围错误

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1e3 + 2e4;const int M = 1040000;struct Edge{    int u;    int v;    int w;}edge[M];int tp, f[N];char st[N][14];bool cmp(struct Edge a, struct Edge b){    return a.w < b.w;}void Init(int n);int get_f(int v);bool Merge(int u, int v);int main(){    int n, m, w, i, j, k;    while(~scanf("%d %d %d", &n, &m, &w)){        getchar();        tp = 0;        for(i = 0; i < n; i++)            scanf("%s", st[i]);        for(i = 0; i < n; i++){            for(j = i+1; j < n; j++){                int cnt = 0;                for(k = 0; k < m; k++){                    if(st[i][k] != st[j][k])                        cnt++;                }                int t;                t = min(m, cnt*w);                edge[tp].u = i, edge[tp].v = j, edge[tp].w = t;                tp++;            }        }        sort(edge, edge+tp, cmp);        Init(n);        int ans = 0, sum = 0;        for(i = 0; i < tp; i++){            if(Merge(edge[i].u, edge[i].v)){                sum += edge[i].w;                ans++;            }            if(ans == n-1)                break;        }        sum += m;        printf("%d\n", sum);    }    return 0;}void Init(int n){    for(int i = 1; i <= n; i++)        f[i] = i;}int get_f(int v){    if(f[v] == v)        return f[v];    f[v] = get_f(f[v]);    return f[v];}bool Merge(int u, int v){    int t1 = get_f(u);    int t2 = get_f(v);    if(t1 == t2)        return false;    else {        f[t2] = t1;        return true;    }}

以下为Accepted代码——Kruskal算法_156ms

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int N = 1e3 + 2e4;const int M = 1040000;struct Edge{    int u;    int v;    int w;}edge[M];int tp, f[N];char st[N][14];bool cmp(struct Edge a, struct Edge b){    return a.w < b.w;}void Init(int n);int get_f(int v);bool Merge(int u, int v);int main(){    int n, m, w, i, j, k;    while(~scanf("%d %d %d", &n, &m, &w)){        getchar();        tp = 0;        for(i = 1; i <= n; i++)            scanf("%s", st[i]);        for(i = 1; i <= n; i++){            for(j = i+1; j <= n; j++){                int cnt = 0;                for(k = 0; k < m; k++){                    if(st[i][k] != st[j][k])                        cnt++;                }                int t;                t = min(m, cnt*w);                edge[tp].u = i, edge[tp].v = j, edge[tp].w = t;                tp++;            }        }        sort(edge, edge+tp, cmp);        Init(n);        int ans = 0, sum = 0;        for(i = 0; i < tp; i++){            if(Merge(edge[i].u, edge[i].v)){                sum += edge[i].w;                ans++;            }            if(ans == n-1)                break;        }        sum += m;        printf("%d\n", sum);    }    return 0;}void Init(int n){    for(int i = 1; i <= n; i++)        f[i] = i;}int get_f(int v){    if(f[v] == v)        return f[v];    f[v] = get_f(f[v]);    return f[v];}bool Merge(int u, int v){    int t1 = get_f(u);    int t2 = get_f(v);    if(t1 == t2)        return false;    else {        f[t2] = t1;        return true;    }}

以下为Accepted代码——Prim算法_15ms

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;const int inf = 0x3f3f3f3f;const int N = 1e3 + 4e2;int vis[N], dis[N], e[N][N];char st[N][14];void Prim(int n, int m);int main(){    int n, m, w, i, j, k;    while(~scanf("%d %d %d", &n, &m, &w)){        getchar();        for(i = 1; i <= n; i++)            scanf("%s", st[i]);        for(i = 1; i <= n; i++){            for(j = i; j <= n; j++){                if(i == j)                    e[i][j] = inf;                else {                    int cnt = 0;                    for(k = 0; k < m; k++){                        if(st[i][k] != st[j][k])                            cnt++;                    }                    int t = min(m, cnt*w);                    e[i][j] = e[j][i] = t;                }            }        }        Prim(n, m);    }    return 0;}void Prim(int n, int m){    int i, num, miv, v, sum;    memset(vis, 0, sizeof(vis));    for(i = 1; i <= n; i++)        dis[i] = e[1][i];    vis[1] = 1, num = 1, sum = 0;    while(num < n){        miv = inf;        for(i = 1; i <= n; i++){            if(!vis[i] && miv > dis[i]){                miv = dis[i];                v = i;            }        }        vis[v] = 1, sum += dis[v], num++;        for(i = 1; i <= n; i++){            if(!vis[i] && dis[i] > e[v][i])                dis[i] = e[v][i];        }    }    printf("%d\n", sum+m);}