[kuangbin带你飞]专题五 【并查集】 【--完结--】
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A - Wireless Network
POJ - 2236AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn = 1005;int n, d;int rep[maxn], x[maxn], y[maxn], f[maxn];void init(){ for(int i = 1; i <= n; ++i) { f[i] = i; rep[i] = 0; }}int getf(int v){ if(f[v] != v) f[v] = getf(f[v]); return f[v];}int dis(int u, int v){ int t1 = x[u] - x[v]; int t2 = y[u] - y[v]; if(t1*t1+t2*t2 <= d*d) return 1; return 0;}int main(){ char ch; int tem, tem1, tem2; int k1, k2; scanf("%d%d",&n, &d); init(); for(int i = 1; i <= n; ++i) scanf("%d%d",&x[i],&y[i]); while(cin >> ch) { if(ch == 'O') { scanf("%d",&tem); rep[tem] = 1; //该电脑被修好，选择能与它联通的电脑加入集合 for(int i = 1; i <= n; ++i) { if(i != tem && rep[i] && dis(tem, i)) { k1 = getf(tem); k2 = getf(i); f[k1] = k2; } } } if(ch == 'S') { scanf("%d%d",&tem1,&tem2); k1 = getf(tem1); k2 = getf(tem2); //cout << k1 << endl; //cout << k2 << endl; if(k1 == k2) cout << "SUCCESS" << endl; else cout << "FAIL" << endl; } } return 0;}B - The Suspects
POJ - 1611AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int MAX = 30005;int f[MAX], num[MAX];int ran[MAX];//树的高度int n, m;void init(){ for(int i = 0; i < n; ++i) { f[i] = i; num[i] = 1; }}int getf(int v){ if(f[v] != v) f[v] = getf(f[v]); return f[v];}void merg(int u, int v){ int t1 = getf(u); int t2 = getf(v); if(t1 == t2) return ; if(ran[t1] > ran[t2]) { //向高的树合并 f[t2] = t1; num[t1] += num[t2]; } else { //树的高度相同时以t2为根进行合并 if(ran[t1] == ran[t2]) ran[t2]++; f[t1]= t2; num[t2] += num[t1]; }}int main(){ int t, tem, tem2; while(scanf("%d%d",&n,&m) && (n||m)) { init(); memset(ran, 0, sizeof(ran)); while(m--) { scanf("%d",&t); scanf("%d",&tem); t--; while(t--) { scanf("%d",&tem2); merg(tem, tem2); } } int ans = getf(0); cout << num[ans] << endl; } return 0;}C - How Many Tables
HDU - 1213AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn = 1005;int n, m;int f[maxn];void init() //注意调用函数init()前先给n赋值{ for(int i = 1; i <= n; ++i) f[i] = i;}int getf(int v){ if(f[v] != v) f[v] = getf(f[v]); return f[v];}void merg(int u, int v){ int t1 = getf(u); int t2 = getf(v); if(t1 != t2) f[t1] = t2;}int main(){ int t, ans, a, b; while(~scanf("%d",&t)) { while(t--) { scanf("%d%d",&n,&m); init(); ans = 0; while(m--) { scanf("%d%d",&a,&b); merg(a, b); } for(int i = 1; i <= n; ++i) if(f[i] == i) ans++; cout << ans << endl; } } return 0;}D - How Many Answers Are Wrong
HDU - 3038AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int N = 200005;int f[N], sum[N];int n, m;int getf(int v){ if(f[v] == v) return v; int t = f[v]; f[v] = getf(f[v]); sum[v] += sum[t]; return f[v];}void merg(int x, int y, int u, int v, int w){ if(x > y) { f[y] = x; sum[y] = sum[u] - sum[v] - w; } else { f[x] = y; sum[x] = w + sum[v] - sum[u]; }}void init(){ for(int i = 0; i <= n; ++i) { f[i] = i; sum[i] = 0; }}int main(){ while(~scanf("%d%d",&n,&m)) { init(); int u, v, w, ans = 0; while(m--) { scanf("%d%d%d",&u,&v,&w); u--; int x = getf(u); int y = getf(v); if(x == y && sum[u] != sum[v] + w) ans++; else if(x != y) merg(x, y, u, v, w); } cout << ans << endl; } return 0;}E - 食物链
POJ - 1182AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn = 100005;int f[150000];int x[maxn], y[maxn], t[maxn];int n, k;void init(int n){ for(int i = 0; i < n; ++i) f[i] = i;}int getf(int v){ if(f[v] != v) f[v] = getf(f[v]); return f[v];}void merg(int u, int v){ int t1 = getf(u); int t2 = getf(v); if(t1 != t2) f[t1] = t2;}int same(int u, int v){ return getf(u) == getf(v);}int main(){ scanf("%d%d",&n,&k); memset(x, 0, sizeof(x)); memset(y, 0, sizeof(y)); memset(t, 0, sizeof(t)); for(int i = 0; i < k; ++i) scanf("%d%d%d",&t[i],&x[i],&y[i]); init(3 * n); //元素x,x+n,x+2*n分别表示x-A,x-B,x-C int ans = 0; for(int i = 0; i < k; ++i) { int tt = t[i]; //物种间的关系 int xx = x[i] - 1; int yy = y[i] - 1; if(xx < 0 || xx >= n || yy < 0 || yy >= n) //假话 { ans++; continue; } if(tt == 1) //同类 { if(same(xx, yy+n) || same(xx, yy+2*n)) ans++; else { merg(xx, yy); merg(xx+n, yy+n); merg(xx+2*n, yy+2*n); } } else //捕食 { if(same(xx, yy) || same(xx, yy+2*n)) ans++; else { merg(xx, yy+n); merg(xx+n, yy+2*n); merg(xx+2*n, yy); } } } cout << ans << endl; return 0;}F - True Liars
POJ - 1417
※AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int MAX = 1000;int f[MAX];int ran[MAX];int sum[2][MAX];int DP[MAX][MAX];int vis[MAX][MAX];void init(){ memset(ran, 0, sizeof(ran)); for(int i = 0; i < MAX; ++i) { f[i] = i; sum[0][i] = 1; sum[1][i] = 0; }}int getf(int v){ if(f[v] == v) return v; int tem = f[v]; f[v] = getf(f[v]); ran[v] = ran[v] ^ ran[tem]; return f[v];}void merg(int u, int v, int k){ int t1 = getf(u); int t2 = getf(v); if(t1 == t2) return ; f[t2] = t1; ran[t2] = k ^ ran[u] ^ ran[v]; sum[0][t1] += sum[0^ran[t2]][t2]; sum[1][t1] += sum[1^ran[t2]][t2];}int main(){ int n, p1, p2; int x, y; string s; while(~scanf("%d%d%d",&n,&p1,&p2) && (n||p1||p2)) { init(); while(n--) { scanf("%d%d",&x,&y); cin >> s; if(s == "yes") merg(x, y, 0); else merg(x, y, 1); } int w1[MAX], w2[MAX], p[MAX]; memset(w1, 0, sizeof(w1)); memset(w2, 0, sizeof(w2)); int cou = 1; int suma = p1 + p2; for(int i = 1; i <= suma; ++i) { if(i == getf(i)) { w1[cou] = sum[0][i]; w2[cou] = sum[1][i]; p[cou] = i; cou++; } } memset(vis, 0, sizeof(vis)); memset(DP, 0, sizeof(DP)); DP[0][0] = 1; for(int i = 1; i < cou; ++i) { for(int t = p1; t >= w1[i]; --t) { if(DP[i-1][t-w1[i]]) { DP[i][t] += DP[i-1][t-w1[i]]; vis[i][t] = 0; } } for(int t = p1; t >= w2[i]; --t) { if(DP[i-1][t-w2[i]]) { DP[i][t] += DP[i-1][t-w2[i]]; vis[i][t] = 1; } } } //如果不能取到或者取到的策略不止一种，不能划分 if(DP[cou-1][p1] != 1) cout << "no" << endl; else { int ans[MAX]; int tot = 0; cou--; while(cou > 0) { int cur = vis[cou][p1]; int roo = p[cou]; for(int i = 1; i <= suma; ++i) { if(getf(i) == roo && ran[i] == cur) ans[tot++] = i; } if(!cur) p1 -= w1[cou]; else p1 -= w2[cou]; cou--; } sort(ans, ans + tot); for(int i = 0; i < tot; ++i) cout << ans[i] << endl; cout << "end" << endl; } } return 0;}G - Supermarket
POJ - 1456AC代码1(比着之前做过的一道题<H>的代码改了改就交了，未用到并查集，不过时间复杂度好像有点高，157MS)：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int maxn = 10005;int a[10005];struct node{ int sco, ded;};int cmp(node x, node y){ if(x.sco == y.sco) return x.ded < y.ded; return x.sco > y.sco;}int main(){ int t, n; while(~scanf("%d",&n)) { long long sum = 0; node f[maxn]; for(int i = 0; i < n; ++i) { scanf("%d",&f[i].sco); scanf("%d",&f[i].ded); sum += f[i].sco; } sort(f, f+n, cmp); long long ans = 0; memset(a, 0, sizeof(a)); for(int i = 0; i < n; ++i) { int tem; for(tem = f[i].ded; tem > 0; --tem) { if(a[tem] == 0) { a[tem] = 1; break; } } if(tem == 0) ans += f[i].sco; } cout << sum - ans << endl; } return 0;}AC代码(并查集，63MS)：
/*************************************************************************主要思想：1.按照价值问题进行排列(贪心2.回溯找值//每次回溯，找前面如果有一个点f[i]==i，那么此点可以使用,就可以停止回溯//此外此点自己--,表示该时间也被使用//后面如果回溯到该点的时候不可以停止下来,仍然需要向前回溯//如果一直回溯到0，那么就是没有位置可以进行//那么这次的价值也不会算到总和里面*************************************************************************/#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int MAX = 1e4 + 5;int f[MAX];struct node{ int value; int time;};bool cmp(node a, node b){ if(a.value == b.value) return a.time > b.time; return a.value > b.value;}void init(){ for(int i = 0; i < MAX; ++i) f[i] = i;}int getf(int v){ if(f[v] != v) f[v] = getf(f[v]); return f[v];}int main(){ int n; while(~scanf("%d",&n)) { init(); node que[MAX]; for(int i = 0; i < n; ++i) { scanf("%d%d",&que[i].value,&que[i].time); } sort(que, que + n, cmp); int ans = 0; for(int i = 0; i < n; ++i) { int tem = que[i].time; int xx = getf(tem); if(xx > 0) { f[xx]--; ans += que[i].value; } } cout << ans << endl; } return 0;}H - Parity game
POJ - 1733
※AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int N = 5005;int a[N*2], n, q, cnt;int f[N*2], r[N*2];struct node{ int u, v; char str[10];} que[N];void init(){ for(int i = 0; i < cnt; ++i) { f[i] = i; r[i] = 0; }}int getf(int v){ if(v != f[v]) { int tem = f[v]; f[v] = getf(f[v]); r[v] = r[v] ^ r[tem]; } return f[v];}int bin(int v){ int low = 0; int high = cnt - 1; int mid; while(low <= high) { mid = (low + high) / 2; if(a[mid] == v) return mid; if(a[mid] < v) low = mid + 1; else high = mid - 1; } return -1;}int main(){ while(~scanf("%d",&n)) { scanf("%d",&q); cnt = 0; for(int i = 0; i < q; ++i) { //注意输入的是数字 数字 字符串 scanf("%d%d%s",&que[i].u,&que[i].v,&que[i].str); que[i].u--; a[cnt++] = que[i].v; a[cnt++] = que[i].u; } sort(a, a+cnt); cnt = unique(a, a+cnt) - a; init(); int ans = 0; for(int i = 0; i < q; ++i) { //这一块按照习惯可以改为一个merg函数 int u = bin(que[i].u); int v = bin(que[i].v); int t1 = getf(u); int t2 = getf(v); if(t1 == t2) { if(r[u] == r[v] && que[i].str[0] == 'o') break; if(r[u] != r[v] && que[i].str[0] == 'e') break; ans++; } else { if(que[i].str[0] == 'o') { f[t1] = t2; r[t1] = r[u] ^ r[v] ^ 1; } else { f[t1] = t2; r[t1] = r[u] ^ r[v]; } ans++; } } cout << ans << endl; } return 0;}I - Navigation Nightmare
POJ - 1984
※AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int MAX = 4e4 + 5;const int MMM = 1e4 + 5;int n, m, k;int f[MAX], ans[MMM];int x[MAX], y[MAX]; //记录坐标int dx[MAX], dy[MAX]; //方向(相对位置)int rx[MAX], ry[MAX]; //最后位置struct node //记录要查询的内容{ int x, y, idx; //x,y是坐标，idx是询问的时间 int n; //记录出现的顺序，方便最后输出}F[MMM];//按照询问时间的先后顺序从小到大排序bool cmp(node u, node v){ return u.idx < v.idx;}void init(){ for(int i = 0; i <= n; ++i) { f[i] = i; rx[i] = ry[i] = 0; }}int getf(int v){ if(f[v] == v) return v; int tem = f[v]; f[v] = getf(f[v]); rx[v] += rx[tem]; ry[v] += ry[tem]; return f[v];}void merg(int u, int v, int tt){ int t1 = getf(u); int t2 = getf(v); f[t2] = t1; rx[t2] = rx[u] - rx[v] - dx[tt]; ry[t2] = ry[u] - ry[v] - dy[tt]; return ;}int main(){ int w, T; char ch; scanf("%d%d",&n,&m); init(); for(int i = 1; i <= m; ++i) { scanf("%d%d%d %c",&x[i], &y[i], &w, &ch); switch(ch) { case 'E': dx[i] = w; dy[i] = 0; break; case 'W': dx[i] = -w; dy[i] = 0; break; case 'N': dx[i] = 0; dy[i] = w; break; case 'S': dx[i] = 0; dy[i] = -w; break; } } scanf("%d",&T); for(int i = 1; i <= T; ++i) { scanf("%d%d%d",&F[i].x,&F[i].y,&F[i].idx); F[i].n = i; } sort(F+1, F+T+1, cmp); //按照询问的先后进行排序 for(int i = 1; i <= T; ++i) { //不断加入点的集合，截止到第F[i].idx次 for(int j = F[i-1].idx; j <= F[i].idx; ++j) { merg(x[j], y[j], j); } if(getf(F[i].x) != getf(F[i].y)) //不连通 ans[F[i].n] = -1; else //求出他们的曼哈顿距离 ans[F[i].n] = abs(rx[F[i].x] - rx[F[i].y]) + abs(ry[F[i].x] - ry[F[i].y]); } for(int i = 1; i <= T; ++i) cout << ans[i] << endl; return 0;}J - A Bug's Life
POJ - 2492
AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int MAX = 4005;int f[MAX];int n, m;void init(int n){ for(int i = 1; i <= n; ++i) f[i] = i;}int getf(int v){ if(f[v] != v) f[v] = getf(f[v]); return f[v];}void merg(int u, int v){ int t1 = getf(u); int t2 = getf(v); if(t1 != t2) f[t1] = t2;}bool same(int u, int v){ return getf(u) == getf(v);}int main(){ int T; bool flag; int x, y; scanf("%d",&T); for(int i = 1; i <= T; ++i) { flag = true; printf("Scenario #%d:\n", i); scanf("%d%d",&n,&m); init(2*n); while(m--) { scanf("%d%d",&x, &y); if(same(x, y)) { flag = false; } else { merg(x, y+n); merg(y, x+n); } } if(flag) printf("No suspicious bugs found!\n"); else printf("Suspicious bugs found!\n"); //if(i != 1) 本来是想这个防PE的，结果用了这个就PE了 cout << endl; } return 0;}K - Rochambeau
POJ - 2912
※AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int N = 505;const int M = 2005;int n, m;bool flag;int f[N], r[N], error[N];struct Node{ int u, v, k;} a[M];void init(){ for(int i = 0; i < n; ++i) { f[i] = i; r[i] = 0; }}int getf(int v){ if(v != f[v]) { int tem = f[v]; f[v] = getf(f[v]); r[v] = (r[v] + r[tem]) % 3; } return f[v];}//有错误&&还没找出来(懒得找了/*void merg(int i, int j, int x, int y, int z){ if(i == x || i == y) return ; int t1 = getf(x); int t2 = getf(y); if(t1 == t2) { if(z==0 && r[x] != r[y]) { error[i] = j + 1; flag = true; return ; } if(z==1 && (r[x]+1)%3 != r[y]) { error[i] = j + 1; flag = true; return ; } if(z==2 && (r[x]+2)%3 != r[y]) { error[i] = j + 1; flag = true; return ; } } else { f[t2] = t1; r[t2] = (r[x] - r[y] + 3 + z + r[t1]) % 3; }}*/int main(){ char ch; while(~scanf("%d%d",&n,&m)) { flag = false; for(int i=0; i<m; i++) { scanf("%d",&a[i].u); char ch; while((ch=getchar())==' '); scanf("%d",&a[i].v); a[i].k=(ch=='=')?0:(ch=='<'?1:2); } memset(error, -1, sizeof(error)); for(int i = 0; i < n; ++i) { init(); for(int j = 0; j < m; ++j) { /*merg(i, j, a[j].u, a[j].v, a[j].k); if(flag) break;*/ if(i==a[j].u||i==a[j].v) continue; int ra=getf(a[j].u),rb=getf(a[j].v); if(ra==rb) { if(a[j].k==0&&r[a[j].u]!=r[a[j].v]) { error[i]=j+1; break; } if(a[j].k==1&&(r[a[j].u]+1)%3!=r[a[j].v]) { error[i]=j+1; break; } if(a[j].k==2&&(r[a[j].u]+2)%3!=r[a[j].v]) { error[i]=j+1; break; } } else { f[rb]=ra; r[rb]=(r[a[j].u]-r[a[j].v]+3+a[j].k+r[ra])%3; } } } int cou, a1, a2; cou = a1 = a2 = 0; for(int i = 0; i < n; ++i) { if(error[i] == -1) { cou++; a1 = i; } a2 = max(a2, error[i]); } if(cou == 0) puts("Impossible"); else if(cou > 1) puts("Can not determine"); else printf("Player %d can be determined to be the judge after %d lines\n",a1,a2); } return 0;}L - Connections in Galaxy War
ZOJ - 3261※AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>#include <map>#include <vector>using namespace std;const int M = 5e4 + 5;const int N = 1e4 + 5;struct node{ int a, b, k;} que[M];map<pair<int, int>, int> edge;int f[N], num[N], val[N];int n, m, q;void init(){ for(int i = 0; i < n; ++i) { f[i] = i; num[i] = val[i]; }}int getf(int v){ if(v != f[v]) { int tem = f[v]; f[v] = getf(f[v]); num[v] = max(num[v], num[tem]); } return f[v];}void merg(int u, int v){ int t1 = getf(u); int t2 = getf(v); if(t1 == t2) return ; if(num[t1] > num[t2] || (num[t1] == num[t2] && t1 < t2)) f[t2] = t1; else f[t1] = t2;}int main(){ bool flag = false; while(~scanf("%d",&n)) { if(flag) cout << endl; else flag = true; for(int i = 0; i < n; ++i) scanf("%d",&val[i]); edge.clear(); scanf("%d",&m); init(); for(int i = 0; i < m; ++i) //创建道路 { int u, v; scanf("%d%d",&u,&v); if(u > v) swap(u, v); edge[make_pair(u, v)] = 1; } scanf("%d",&q); for(int i = 0; i < q; ++i) { char str[10]; scanf("%s",str); if(str[0] == 'q') //查询 { scanf("%d",&que[i].a); que[i].k = 0; } else //摧毁道路 { scanf("%d%d",&que[i].a,&que[i].b); if(que[i].a > que[i].b) swap(que[i].a, que[i].b); edge[make_pair(que[i].a, que[i].b)] = 0; que[i].k = 1; } } map<pair<int, int>, int> :: iterator it; for(it = edge.begin(); it != edge.end(); ++it) { if(it -> second) { pair<int, int> tem = it -> first; merg(tem.first, tem.second); } } vector<int> ans; ans.clear(); for(int i = q-1; i >= 0; --i) { if(que[i].k == 0) { if(num[getf(que[i].a)] <= val[que[i].a]) ans.push_back(-1); else ans.push_back(getf(que[i].a)); } else merg(que[i].a, que[i].b); } for(int i = ans.size()-1; i >= 0; --i) cout << ans[i] << "\n"; } return 0;}M - 小希的迷宫
HDU - 1272
AC代码：
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int MAX = 1e5 + 5;int f[MAX];bool vis[MAX];int n, m;void init(){ for(int i = 0; i < MAX; ++i) { f[i] = i; vis[i] = false; }}int getf(int v){ if(f[v] != v) f[v] = getf(f[v]); return f[v];}bool merg(int u, int v){ int t1 = getf(u); int t2 = getf(v); if(t1 != t2) { f[t1] = t2; return true; } return false; //两点的根节点一样——成环}int main(){ while(scanf("%d%d",&n,&m) && (n != -1 || m != -1)) { if(!n && !m) //如果第一组就是0 0 { cout << "Yes" << endl; continue; } init(); //开始建树 vis[n] = vis[m] = true; bool flag = merg(n, m); while(scanf("%d%d",&n,&m) && (n || m)) { vis[n] = vis[m] = true; if(!merg(n, m)) flag = false; } int cou = 0; if(!flag) { cout << "No" << endl; continue; } else { for(int i = 0; i < MAX; ++i) if(vis[i] && f[i] == i) cou++; } if(cou > 1) { cout << "No" << endl; continue; } cout << "Yes" << endl; } return 0;}N - Is It A Tree?
POJ - 1308
#include <cstdio>#include <iostream>#include <cstring>#include <algorithm>#include <cmath>using namespace std;const int MAX = 1e5 + 5;int f[MAX];bool vis[MAX];int n, m;void init(){ for(int i = 0; i < MAX; ++i) { f[i] = i; vis[i] = false; }}int getf(int v){ if(f[v] != v) f[v] = getf(f[v]); return f[v];}bool merg(int u, int v){ int t1 = getf(u); int t2 = getf(v); if(t1 != t2) { f[t1] = t2; return true; } return false; //两点的根节点一样——成环}int main(){ int cas = 0; while(scanf("%d%d",&n,&m) && (n != -1 || m != -1)) { if(!n && !m) //如果第一组就是0 0 { printf("Case %d is a tree.\n", ++cas); continue; } init(); //开始建树 vis[n] = vis[m] = true; bool flag = merg(n, m); while(scanf("%d%d",&n,&m) && (n || m)) { vis[n] = vis[m] = true; if(!merg(n, m)) flag = false; } int cou = 0; if(!flag) { printf("Case %d is not a tree.\n", ++cas); continue; } else { for(int i = 0; i < MAX; ++i) if(vis[i] && f[i] == i) cou++; } if(cou > 1) { printf("Case %d is not a tree.\n", ++cas); continue; } printf("Case %d is a tree.\n", ++cas); } return 0;} 0 0
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