{ int v = 0; int pathsum=0; int max= inf; int minpath; int vis[12]; int j=0; memset(vis,0,sizeof(vis)); vis[0] = 1; cout<<"(0)公园正门"; for(int flat=0; flat!=n;) { minpath=inf; for(int k=0; k<8; k++) { if(minpath>path[v][k] && vis[k]==0) { minpath=path[v][k]; j=k; } } if(minpath<max) { flat++; vis[j]=1; pathsum=pathsum+minpath; v=j; cout<<"->("<<v<<")"<<name[v].na; } if(flat==n-1) vis[0]=0; if(flat==n) cout<<endl<<"全程"<<pathsum<<"公里"<<endl; } }这段代码是什么功能
时间: 2024-04-25 08:26:10 浏览: 47
这段代码是一段实现最小生成树算法的程序,用于求解一个带权无向图的最小生成树。其中,输入的图有8个节点,通过计算各节点之间的路径长度,找到一条最短路径,并输出最小生成树的总长度和路径。其中,变量v表示当前节点,变量pathsum表示路径长度之和,变量vis表示节点是否被访问过的标记,变量j表示下一个节点,变量name表示节点的名称,变量inf表示一个非常大的数字,用于初始化最小路径长度。
相关问题
{ int v = 0; int pathsum=0; int max= inf; int minpath; int vis[12]; int j=0; memset(vis,0,sizeof(vis)); vis[0] = 1; cout<<"(0)公园正门"; for(int flat=0; flat!=n;) { minpath=inf; for(int k=0; k<8; k++) { if(minpath>path[v][k] && vis[k]==0) { minpath=path[v][k]; j=k; } }
这段代码是一个基于贪心算法的最短路算法,用于找到从公园正门出发经过所有景点且总路程最短的路径。其中,变量v表示当前所在的景点,pathsum表示当前路径的总路程,max表示路径上最大的边权值(用于剪枝),minpath表示当前可选的最短路径长度,vis数组用于记录每个景点是否已经被访问过,j表示下一个要访问的景点的标号。在循环中,首先找到当前可选路径中的最短路径,然后将下一个要访问的景点标记为已访问,并更新当前路径的总路程和当前所在景点的标号。当经过景点的数量达到n-1时(即除了起点公园正门外,已经经过了所有的景点),将公园正门的状态改为未访问,以便找到一条回到公园正门的最短路径。当经过景点的数量达到n时(即已经回到公园正门),输出总路程信息。
#include <algorithm> #include <cstdio> #include <map> #include <queue> using namespace std; const int maxn = 205; const int INF = 0x3f3f3f3f; int d[maxn][maxn]; int terminal[maxn], vis[maxn][maxn]; map<int, int> been[maxn]; int n, m, k; int line[10000]; int main() { scanf("%d%d%d", &n, &m, &k); for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) d[i][j] = (i == j) ? 0 : INF; int u, v, len; int fare; char ch; while (m--) { int len = 0; while (scanf("%d", &u)) { line[len++] = u; ch = getchar(); if (ch == '\n') { terminal[line[0]] = terminal[line[len - 1]] = 1; for (int i = 0; i != len - 1; i += 2) { u = line[i], v = line[i + 2]; d[v][u] = d[u][v] = min(d[u][v], line[i + 1]); } break; } } } for (int k = 1; k <= n; k++) { for (int i = 1; i <= n; i++) for (int j = 1; j <= n; j++) d[i][j] = min(d[i][j], d[i][k] + d[k][j]); } for (int i = 1; i <= n; i++) { for (int j = 1; j <= n; j++) { if (i == j || d[i][j] == INF) continue; fare = 2 + d[i][j] / k; if (!been[i].count(fare) || been[i][fare] < d[i][j]) been[i][fare] = d[i][j]; } } int t, cur, first; queue<int> Q; scanf("%d", &t); while (t--) { first = 1; scanf("%d", &u); vis[u][u] = 1; Q.push(u); while (!Q.empty()) { cur = Q.front(); Q.pop(); for (int i = 1; i <= n; i++) { if (vis[u][i] || d[cur][i] == INF) continue; if (terminal[i]) { Q.push(i); vis[u][i] = 1; } else { fare = 2 + d[cur][i] / k; if (d[cur][i] == been[cur][fare]) { Q.push(i); vis[u][i] = 1; } } } } for (int i = 1; i <= n; i++) { if (vis[u][i]) { if (first) { printf("%d", i); first = 0; } else printf(" %d", i); } } printf("\n"); } return 0; }把这段代码改为C语言代码
下面是将该段代码改为C语言的代码:
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#define maxn 205
#define INF INT_MAX
int d[maxn][maxn];
int terminal[maxn], vis[maxn][maxn];
int been[maxn][maxn];
int n, m, k;
int line[10000];
int main() {
scanf("%d%d%d", &n, &m, &k);
// 初始化距离矩阵
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
d[i][j] = (i == j) ? 0 : INF;
}
}
// 读入边并建图
int u, v, len, fare;
char ch;
for (int i = 0; i < m; i++) {
len = 0;
while (scanf("%d", &u)) {
line[len++] = u;
ch = getchar();
if (ch == '\n') {
terminal[line[0]] = terminal[line[len - 1]] = 1;
for (int i = 0; i != len - 1; i += 2) {
u = line[i], v = line[i + 2];
d[v][u] = d[u][v] = (d[u][v] > line[i + 1]) ? line[i + 1] : d[u][v];
}
break;
}
}
}
// Floyd算法求最短路
for (int k = 1; k <= n; k++) {
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (d[i][k] != INF && d[k][j] != INF && d[i][j] > d[i][k] + d[k][j]) {
d[i][j] = d[i][k] + d[k][j];
}
}
}
}
// 记录每个站点到其他站点的最小票价
for (int i = 1; i <= n; i++) {
for (int j = 1; j <= n; j++) {
if (i == j || d[i][j] == INF) {
continue;
}
fare = 2 + d[i][j] / k;
if (been[i][fare] < d[i][j]) {
been[i][fare] = d[i][j];
}
}
}
// 处理查询
int t, cur, first;
scanf("%d", &t);
while (t--) {
first = 1;
scanf("%d", &u);
vis[u][u] = 1;
for (int i = 1; i <= n; i++) {
if (terminal[i]) {
vis[u][i] = 1;
} else {
fare = 2 + d[u][i] / k;
if (d[u][i] == been[i][fare]) {
vis[u][i] = 1;
}
}
}
for (int i = 1; i <= n; i++) {
if (vis[u][i]) {
if (first) {
printf("%d", i);
first = 0;
} else {
printf(" %d", i);
}
}
}
printf("\n");
}
return 0;
}
```