URAL 1003,1004
1003:
并查集在处理矛盾关系的应用，讲的比较好的题解
#include <map> #include <set> #include <list> #include <queue> #include <stack> #include <cmath> #include <ctime> #include <vector> #include <bitset> #include <cstdio> #include <string> #include <numeric> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #include <functional> using namespace std; typedef long long ll; typedef unsigned long long ull; int dx[4]={-1,1,0,0}; int dy[4]={0,0,-1,1};//up down left right bool inmap(int x,int y,int n,int m){if(x<1||x>n||y<1||y>m)return false;return true;} int hashmap(int x,int y,int m){return (x-1)*m+y;} #define eps 1e-8 #define inf 0x7fffffff #define debug puts("BUG") #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define read freopen("in.txt","r",stdin) #define write freopen("out.txt","w",stdout) #define N 11111 map<int,int>mp; struct node { int l,r; char c; }nd[N>>1]; int fa[N<<1]; int find(int x) { if (fa[x]!=x) fa[x] = find(fa[x]); return fa[x]; } int gao(int m,int cnt) { for (int i = 0; i < m; ++i) { int l = mp[nd[i].l-1], r= mp[nd[i].r]; char c = nd[i].c; int f1 = find(l), f2 = find(r), f3 = find(l+cnt), f4 = find(r+cnt); if (c == 'e') { if (f1 == f4 && f2 == f3) return i; fa[f1] = f2; fa[f3] = f4; } else { if (f1 == f2) return i; fa[f1] = f4; fa[f2] = f3; } } return m; } int main() { //read; int n,m; while (~scanf("%d",&n)) { if (n == -1) break; scanf("%d",&m); mp.clear(); int l, r, cnt = 0; char str[11]; for (int i = 0; i < m; ++i) { scanf("%d%d%s",&l,&r,str); nd[i].l = l, nd[i].r = r; nd[i].c = str[0]; if (mp.find(l-1) == mp.end()) mp[l-1] = cnt++; if (mp.find(r) == mp.end()) mp[r] = cnt++; } for (int i = 0; i < 2*cnt; ++i) fa[i] = i; int ans = gao(m,cnt); printf("%d\n", ans); } return 0; }
1004：
能够加深对floyd理解的好题，当然还有另外的做法
#include <map> #include <set> #include <list> #include <queue> #include <stack> #include <cmath> #include <ctime> #include <vector> #include <bitset> #include <cstdio> #include <string> #include <numeric> #include <cstring> #include <cstdlib> #include <iostream> #include <algorithm> #include <functional> using namespace std; typedef long long ll; typedef unsigned long long ull; int dx[4]={-1,1,0,0}; int dy[4]={0,0,-1,1};//up down left right bool inmap(int x,int y,int n,int m){if(x<1||x>n||y<1||y>m)return false;return true;} int hashmap(int x,int y,int m){return (x-1)*m+y;} #define eps 1e-8 #define inf 0x7ffffff #define debug puts("BUG") #define lson l,m,rt<<1 #define rson m+1,r,rt<<1|1 #define read freopen("in.txt","r",stdin) #define write freopen("out.txt","w",stdout) #define N 155 int mp[N][N]; int dis[N][N]; int nxt[N][N]; stack<int>st; int floyd(int n) { int ans = inf; for (int k = 1; k <= n; ++k) { for (int i = 1; i < k; ++i) for (int j = i+1; j < k; ++j) { if (ans > dis[i][j] + mp[k][i] + mp[j][k]) { ans = dis[i][j] + mp[k][i] + mp[j][k]; while (!st.empty()) st.pop(); for (int t = i; t != j; t = nxt[t][j]) st.push(t); st.push(j); st.push(k); } } for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) { if (dis[i][k] == inf || dis[k][j] == inf) continue; if (dis[i][k] + dis[k][j] < dis[i][j]) { dis[i][j] = dis[i][k] + dis[k][j]; nxt[i][j] = nxt[i][k]; } } } return ans; } int main() { //read; int n,m; while (~scanf("%d",&n)) { if (n == -1)break; scanf("%d",&m); for (int i = 1; i <= n; ++i) for (int j = 1; j <= n; ++j) { dis[i][j] = mp[i][j] = inf; nxt[i][j] = j; } int u,v,c; for (int i = 0; i < m; ++i) { scanf("%d%d%d",&u, &v, &c); if (mp[u][v] > c) mp[u][v] = mp[v][u] = dis[u][v] = dis[v][u] = c; } int ans = floyd(n); //printf("%d\n",ans); if (ans == inf) puts("No solution."); else { bool f = false; while (!st.empty()) { if (f) printf(" "); else f = true; printf("%d",st.top()); st.pop(); }puts(""); } } return 0; }
