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BZOJ1455 罗马游戏
支持合并,删除最小值操作的不就是可并堆吗。。。
还要窝有板子,液!
这题还要注意,对于一个点,找根的话,要用到并查集。
1 /************************************************************** 2 Problem: 1455 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:1316 ms 7 Memory:21312 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 using namespace std; 14 const int N = 1000001; 15 int n, m, fa[N]; 16 bool died[N]; 17 18 struct heap{ 19 int v, l, r, dep; 20 }h[N]; 21 22 int Cnt; 23 24 inline int read(){ 25 int x = 0; 26 char ch = getchar(); 27 while (ch < ‘0‘ || ch > ‘9‘) 28 ch = getchar(); 29 while (ch >= ‘0‘ && ch <= ‘9‘){ 30 x = x * 10 + ch - ‘0‘; 31 ch = getchar(); 32 } 33 return x; 34 } 35 36 int pr[10], NUM = 0; 37 inline void print(int x){ 38 while (x) 39 pr[++NUM] = x % 10, x /= 10; 40 while (NUM) 41 putchar(pr[NUM--] + ‘0‘); 42 putchar(‘\n‘); 43 } 44 45 int find_fa(int x){ 46 return fa[x] == x ? x : fa[x] = find_fa(fa[x]); 47 } 48 49 inline int new_heap(int x){ 50 h[++Cnt].v = x; 51 h[Cnt].l = h[Cnt].r = h[Cnt].dep = 0; 52 return Cnt; 53 } 54 55 int Merge(int x, int y){ 56 if (!x || !y) return x + y; 57 if (h[x].v > h[y].v) 58 swap(x, y); 59 h[x].r = Merge(h[x].r, y); 60 if (h[h[x].l].dep < h[h[x].r].dep) 61 swap(h[x].l, h[x].r); 62 h[x].dep = h[h[x].r].dep + 1; 63 return x; 64 } 65 66 int Top(int x){ 67 return h[x].v; 68 } 69 70 int Pop(int x){ 71 return Merge(h[x].l, h[x].r); 72 } 73 74 int main(){ 75 n = read(); 76 int i, t, x, y; 77 for (i = 1; i <= n; ++i) 78 new_heap(read()), fa[i] = i; 79 h[0].dep = -1; 80 char ch; 81 m = read(); 82 for (i = 1; i <= m; ++i){ 83 ch = getchar(); 84 while (ch != ‘M‘ && ch != ‘K‘) ch = getchar(); 85 if (ch == ‘M‘){ 86 x = read(), y = read(); 87 if (died[x] || died[y]) continue; 88 x = find_fa(x), y = find_fa(y); 89 if (x != y) 90 t = Merge(x, y), fa[x] = fa[y] = t; 91 }else{ 92 x = read(); 93 if (died[x]){ 94 puts("0"); 95 continue; 96 } 97 x = find_fa(x); 98 died[x] = 1; 99 print(h[x].v);100 fa[x] = Merge(h[x].l, h[x].r);101 fa[fa[x]] = fa[x];102 }103 }104 return 0;105 }
BZOJ1455 罗马游戏
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