zoj 2750 Idiomatic Phrases Game

迪杰斯特拉单源最短路算法。对成语进行预处理。做出邻接矩阵即可。

#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int maxn = ;
int c[maxn], len[maxn], cost[maxn][maxn], flag[maxn], e[maxn];
char s[maxn][];
int main()
{
    int n, i, j, ii;
    while (~scanf("%d", &n))
    {
        if (n == ) break;
        for (i = ; i < n; i++)    scanf("%d%s", &c[i], s[i]);
        for (i = ; i < n; i++) len[i] = strlen(s[i]);
        memset(flag, , sizeof(flag));
        for (i = ; i <= n; i++) for (j = ; j <= n; j++) cost[i][j] = ;
        for (i = ; i < n; i++)
        {
            for (j = ; j < n; j++)
            {
                if (i != j&&s[j][] == s[i][len[i] - ] && s[j][] == s[i][len[i] - ] && s[j][] == s[i][len[i] - ] && s[j][] == s[i][len[i] - ])
                    cost[i][j] = c[i];
            }
        }
        for (i = ; i < n; i++) e[i] = cost[][i];
        e[] = ; flag[] = ; int x, uu;
        for (ii = ; ii < n - ; ii++)
        {
            int minn = ; uu = ;
            for (i = ; i < n; i++)
            {
                if (!flag[i] && e[i] < minn)
                {
                    x = i;
                    minn = e[i];
                    uu = ;
                }
            }
            if (!uu) continue;
            flag[x] = ;
            for (i = ; i < n; i++)
            if (!flag[i] && cost[x][i] !=  && e[x] + cost[x][i] < e[i])
                e[i] = e[x] + cost[x][i];
        }
        if (e[n - ] != ) printf("%d\n", e[n - ]);
        else printf("-1\n");
    }
    return ;
}