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 = 1005;
int c[maxn], len[maxn], cost[maxn][maxn], flag[maxn], e[maxn];
char s[maxn][100];
int main()
{
    int n, i, j, ii;
    while (~scanf("%d", &n))
    {
        if (n == 0) break;
        for (i = 0; i < n; i++)    scanf("%d%s", &c[i], s[i]);
        for (i = 0; i < n; i++) len[i] = strlen(s[i]);
        memset(flag, 0, sizeof(flag));
        for (i = 0; i <= n; i++) for (j = 0; j <= n; j++) cost[i][j] = 999999999;
        for (i = 0; i < n; i++)
        {
            for (j = 0; j < n; j++)
            {
                if (i != j&&s[j][0] == s[i][len[i] - 4] && s[j][1] == s[i][len[i] - 3] && s[j][2] == s[i][len[i] - 2] && s[j][3] == s[i][len[i] - 1])
                    cost[i][j] = c[i];
            }
        }
        for (i = 0; i < n; i++) e[i] = cost[0][i];
        e[0] = 0; flag[0] = 1; int x, uu;
        for (ii = 0; ii < n - 1; ii++)
        {
            int minn = 999999999; uu = 0;
            for (i = 0; i < n; i++)
            {
                if (!flag[i] && e[i] < minn)
                {
                    x = i;
                    minn = e[i];
                    uu = 1;
                }
            }
            if (!uu) continue;
            flag[x] = 1;
            for (i = 0; i < n; i++)
            if (!flag[i] && cost[x][i] != 999999999 && e[x] + cost[x][i] < e[i])
                e[i] = e[x] + cost[x][i];
        }
        if (e[n - 1] != 999999999) printf("%d\n", e[n - 1]);
        else printf("-1\n");
    }
    return 0;
}