zoj 2027 Travelling Fee

SPFA+枚举。

每条边的权值都设置一次为0 用一次SPFA，算出最短路，每次的最短路取最小值就是答案。

 

#include<stdio.h>

#include<string.h>

#include<math.h>

#include<map>

#include<queue>

#include<vector>

#include<string>

#include<iostream>

#include<algorithm>

using namespace std;



const int maxn = 210;

map<string, int>zh;

vector<int>ljb[maxn];

int cost[maxn][maxn];//邻接矩阵

string s, s1, s2;

int qidian, zhongdian, n, id;

int dist[maxn], flag[maxn];



void spfa()

{

    int iii;

    queue<int>Q;

    memset(flag, 0, sizeof(flag));

    for (iii = 0; iii<id; iii++) dist[iii] = 999999999;

    dist[qidian] = 0; Q.push(qidian); flag[qidian] = 1;

    while (!Q.empty())

    {

        int h = Q.front(); Q.pop(); flag[h] = 0;

        for (iii = 0; iii<ljb[h].size(); iii++)

        {

            if (cost[h][ljb[h][iii]] != 999999999)

            {

                if (dist[h] + cost[h][ljb[h][iii]]<dist[ljb[h][iii]])

                {

                    dist[ljb[h][iii]] = dist[h] + cost[h][ljb[h][iii]];

                    if (flag[ljb[h][iii]] == 0)

                    {

                        Q.push(ljb[h][iii]);

                        flag[ljb[h][iii]] = 1;

                    }

                }

            }

        }

    }

}



int main()

{

    int i, j, cc;

    while (cin >> s)

    {

        for (i = 0; i<210; i++) ljb[i].clear();

        zh.clear();

        id = 1; zh[s] = id; id++;

        qidian = 1; zhongdian = 2; cin >> s;

        zh[s] = id; id++;

        scanf("%d", &n);

        for (i = 0; i <= 205; i++)

        {

            for (j = 0; j <= 205; j++)

            {

                if (i == j) cost[i][j] = 0;

                else cost[i][j] = 999999999;

            }

        }

        for (i = 0; i<n; i++)

        {

            cin >> s1 >> s2 >> cc;

            if (zh[s1] == 0)zh[s1] = id, id++;

            if (zh[s2] == 0)zh[s2] = id, id++;

            cost[zh[s1]][zh[s2]] = cc;

            ljb[zh[s1]].push_back(zh[s2]);

        }

        int anss = 999999999;

        for (i = 1; i<id; i++)

        {

            for (j = 1; j<id; j++)

            {

                if (i != j&&cost[i][j] != 999999999)

                {

                    int t = cost[i][j];

                    cost[i][j] = 0;

                    spfa();

                    if (dist[zhongdian]<anss) anss = dist[zhongdian];

                    cost[i][j] = t;

                }

            }

        }

        printf("%d\n", anss);

    }

    return 0;

}