題意：輸入一個n(n20)個節(jié)點的無向圖以及某個結(jié)點k，按照字典序從小到大順序輸出從結(jié)點1到結(jié)點k的所有路徑，要求經(jīng)過結(jié)點不能重復(fù)經(jīng)過。

分析：終于做到水題了，此題唯一要點就是先判斷是否能到達，用dfs。

#include<cstdio> #include<cstring> #include<cctype> #include<queue> #include<iostream> #include<vector> #include<list> #include<set> using namespace std; const int maxn = 20+5; int G[maxn][maxn], vis[maxn]; int k,m; int allroute; int routemap[400][maxn]; void solve(int cur,int dep) {if (cur == k) {for (int i = 0; i < dep-1; i++) {cout << routemap[allroute][i] << " ";}cout << routemap[allroute][dep - 1];cout << endl;allroute++;memcpy(routemap[allroute], routemap[allroute - 1], sizeof(routemap[allroute - 1]));return;}for (int i = 1; i <= m; i++) {if (G[cur][i]&&!vis[i]) {vis[i] = 1;routemap[allroute][dep] = i;solve(i, dep + 1);vis[i] = 0;}} } bool isarrive(int cur) {if (cur == k) {return true;}for (int i = 1; i <= m; i++) {if (G[cur][i] && !vis[i]) {vis[i] = 1;if (isarrive(i))return true;}}return false; } int main() {int a, b; int kase = 0;while (cin >> k && k) {++kase;memset(G, 0, sizeof(G));memset(vis, 0, sizeof(vis));m = 0; allroute = 0;while (cin >> a >> b && a&&b) {G[a][b] = 1;G[b][a] = 1;m = max(max(a, b), m);}cout << "CASE " << kase << ":" << endl;if (isarrive(1)) {memset(vis, 0, sizeof(vis));vis[1] = 1;routemap[allroute][0] = 1;solve(1, 1);cout << "There are "<<allroute<< " routes from the firestation to streetcorner " << k << "." << endl;}else {//不可達cout << "There are 0 routes from the firestation to streetcorner " << k << "." << endl;}}//system("pause");return 0; }

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