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PAT1087. All Roads Lead to Rome

时间:2017-11-26 16:48:26      阅读:190      评论:0      收藏:0      [点我收藏+]

标签:path   min   name   lead   相同   clu   一个   stream   tor   

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\n", cntRo, dis[rom], ansHapy, (int)ansAvg);
    for(int i = anspath.size() - 1; i != 0; i--){
        cout << i_s[anspath[i]] << "->";
    }
    printf("ROM");
    return 0;
}

PAT1087. All Roads Lead to Rome

标签:path   min   name   lead   相同   clu   一个   stream   tor   

原文地址:http://www.cnblogs.com/1pha/p/7899190.html

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