PTA甲级：

1003 Emergency

经典迪杰斯特拉算法

#include<iostream>
#include<cstring>using namespace std;const int N = 510 , INF = 0x3f3f3f3f;
int dist[N] , cnt[N] , g[N][N];
int helper[N] , way[N];
bool st[N];
int n , m , c1 , c2;int main()
{cin >> n >> m >> c1 >> c2;for(int i = 0;i < n;i ++)cin >> cnt[i];memset(dist , 0x3f , sizeof dist);memset(g , 0x3f , sizeof g);while(m --){int a , b , c;cin >> a >> b >> c;g[a][b] = g[b][a] = c;}dist[c1] = 0 , way[c1] = 1 , helper[c1] = cnt[c1];for(int i = 0;i < n;i ++){int t = -1;for(int j = 0;j < n;j ++)if(!st[j] && (t == -1 || dist[t] > dist[j]))t = j;st[t] = true;for(int j = 0;j < n;j ++){if(dist[j] > dist[t] + g[t][j]){dist[j] = dist[t] + g[t][j];way[j] = way[t];helper[j] = helper[t] + cnt[j];}else if(dist[j] == dist[t] + g[t][j]){way[j] += way[t];helper[j] = max(helper[t] + cnt[j] , helper[j]);}}}cout << way[c2] << " " << helper[c2] << endl;return 0;
}

1004 Counting Leaves

经典深度优先方法dfs

#include<iostream>
#include<vector>using namespace std;const int N = 200;
int n , m;
vector<int>v[N];
int cnt[N];
int dep = 1;void dfs(int x , int depth)
{dep = max(dep , depth);if(!v[x].size()) {cnt[depth] ++;return ;}for(auto i : v[x])dfs(i , depth + 1);
}int main()
{while(cin >> n){if(!n) break;cin >> m;for(int i = 0;i < m;i ++){int id;cin >> id;int k;cin >> k;while(k --){int id1;cin >> id1;v[id].push_back(id1);}}dfs(1 , 1);for(int i = 1;i <= dep;i ++)if(i == 1) cout << cnt[i];else cout << " " << cnt[i];cout << endl;}return 0;
}

天梯赛：

L2-001 紧急救援

同1003一样

L2-048 寻宝图

经典dfs

#include<iostream>
#include<vector>
#include<queue>using namespace std;const int N = 1e5 + 10;
int n , m;
int total , tre;
vector<int>nodes[N];
vector<bool>st[N];
int dx[4] = {0 , 0 , 1 , -1};
int dy[4] = {1 , -1 , 0 , 0};void dfs(int x , int y , bool &f)
{if(nodes[x][y] > 1) f = true;st[x][y] = true;for(int i = 0;i < 4;i ++){int tx = x + dx[i] , ty = y + dy[i];if(tx < 0 || tx >= n || ty < 0 || ty >= m) continue;if(st[tx][ty] || !nodes[tx][ty]) continue;dfs(tx , ty , f);}
}int main()
{cin >> n >> m;for(int i = 0;i < n;i ++){string s;cin >> s;for(int j = 0;j < m;j ++)nodes[i].push_back(s[j] - '0') , st[i].push_back(false);}for(int i = 0;i < n;i ++){for(int j = 0;j < m;j ++){if(nodes[i][j] && !st[i][j]) {bool f = false;dfs(i , j , f);if(f) tre ++;total ++;}}}cout << total << " " << tre << endl;return 0;
}