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;

}