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  • PAT1087. All Roads Lead to Rome

    PAT1087. All Roads Lead to Rome

    题目大意

    给定一个图的边权和点权, 求边权最小的路径; 若边权相同, 求点权最大; 若点权相同, 则求平均点权最大.

    思路

    先通过 Dijkstra 求得最短路径, 需要注意的是: 要保证每次松弛时 u 和 v 不相同, 否则会形成自环, 则从 ROM 开始 BFS 遍历每一条边权相同的路径.

    代码

    #include <iostream>
    #include <cstdio>
    #include <vector>
    #include <map>
    using namespace std;
    #define MAXN 300
    #define INF 0x7ffffff
    int nVertex, nEdge;
    map<string, int> s_i;
    map<int, string> i_s;
    vector<int> prepath[MAXN], temppath, anspath;
    int vw[MAXN];
    int ew[MAXN][MAXN];
    int dis[MAXN];
    int isVis[MAXN];
    int cntRo = 0;
    int ansHapy = 0;
    double ansAvg = 0;
    void dfs(int loc){
        temppath.push_back(loc);
        if(loc == 0){
            int hapy = 0;
            for(int i = 0; i < temppath.size(); i++)
                hapy += vw[temppath[i]];
            double avgHapy = hapy * 1.0 / (temppath.size() - 1);
            if(hapy > ansHapy){
                ansHapy = hapy;
                ansAvg = avgHapy;
                anspath = temppath;
            }
            else if(hapy == ansHapy && avgHapy > ansAvg){
                ansAvg = avgHapy;
                anspath = temppath;
            }
            cntRo++;
            temppath.pop_back();
            return;
        }
        for(int i = 0; i < prepath[loc].size(); i++){
            dfs(prepath[loc][i]);
        }
        temppath.pop_back();
    }
    int main(){
        scanf("%d%d", &nVertex, &nEdge);
        string tempStr; cin >> tempStr;
        s_i[tempStr] = 0; i_s[0] = tempStr;
        for(int i = 1; i < nVertex; i++){
            cin >> tempStr;
            s_i[tempStr] = i; i_s[i] = tempStr;
            scanf("%d", &vw[i]);
        }
        for(int i = 0; i < MAXN; i++){
            for(int j = 0; j < MAXN; j++){
                ew[i][j] = (i == j ? 0 : INF);
            }
        }
        for(int i = 0; i < nEdge; i++){
            string a, b; int c;
            cin >> a >> b >> c;
            ew[s_i[a]][s_i[b]] = ew[s_i[b]][s_i[a]] = c;
        }
    
        for(int i = 0; i < nVertex; i++)
            dis[i] = ew[0][i];
        dis[0] = 0;
        for(int i = 0; i < nVertex; i++){
            int u = -1, minn = INF;
            for(int j = 0; j < nVertex; j++){
                if(!isVis[j] && dis[j] < minn){
                    minn = dis[j];
                    u = j;
                }
            }
            isVis[u] = 1;
            for(int v = 0; v < nVertex; v++){
                if(u != v)
                {
                    if(dis[v] > ew[u][v] + dis[u]){
                        dis[v] = ew[u][v] + dis[u];
                        prepath[v].clear();
                        prepath[v].push_back(u);
                    }
                    else if(dis[v] == ew[u][v] + dis[u]){
                        prepath[v].push_back(u);
                    }
                }
            }
        }
        int rom = s_i["ROM"];
        dfs(rom);
        printf("%d %d %d %d
    ", cntRo, dis[rom], ansHapy, (int)ansAvg);
        for(int i = anspath.size() - 1; i != 0; i--){
            cout << i_s[anspath[i]] << "->";
        }
        printf("ROM");
        return 0;
    }
    
    
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  • 原文地址:https://www.cnblogs.com/1pha/p/7899190.html
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