标签:path min name lead 相同 clu 一个 stream tor
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.
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.
#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