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
", anss);
    }
    return 0;
}