题意：
给出n个点和m条边<边是蓝色或者红色>的信息..问能不能组成一棵有k条蓝边的树..
思路：
求出尽量多地加蓝边可以加多少条..bl
然后求出尽量多地加红边可以加多少条..re
如果k是在bl和re间就是1 否则是0
Tips：
可以用kruskal也可以之间建树..
Code：
View Code
1 #include <stdio.h> 2 #include <cstring> 3 #include <cmath> 4 #include <algorithm> 5 using namespace std; 6 #define clr(x) memset(x, 0, sizeof(x)) 7 int n, m, k; 8 9 int f[10510]; 10 int find(int x) 11 { 12 return f[x] == x?x:f[x] = find(f[x]); 13 } 14 15 struct Edge 16 { 17 int u, v; 18 char col; 19 }edge[1100005]; 20 21 int cmp1(Edge a, Edge b) 22 { 23 return a.col < b.col; 24 } 25 26 int cmp2(Edge a, Edge b) 27 { 28 return a.col > b.col; 29 } 30 31 int solve(bool flag) 32 { 33 int ans = 0; 34 for(int i = 0; i <= n; ++i) 35 f[i] = i; 36 37 if(flag) 38 sort(edge, edge+m, cmp1); 39 else 40 sort(edge, edge+m, cmp2); 41 42 for(int i = 0; i < m; ++i) { 43 int xx = find(edge[i].u), yy = find(edge[i].v); 44 if(xx != yy){ 45 f[xx] = yy; 46 if(edge[i].col == 'B') 47 ans++; 48 } 49 } 50 return ans; 51 } 52 53 int main() 54 { 55 int i, j; 56 int bl, re; 57 while(scanf("%d %d %d", &n, &m, &k) != EOF) 58 { 59 if(n == 0 && m == 0 && k == 0) break; 60 bl = re = 0; 61 62 for(i = 0; i < m; ++i){ 63 getchar(); 64 scanf("%c%d%d", &edge[i].col, &edge[i].u, &edge[i].v); 65 } 66 67 bl = solve(true); 68 re = solve(false); 69 if(k <= bl && k >= re) puts("1"); 70 else puts("0"); 71 } 72 return 0; 73 }
其实不排序都可以..
只要是符合条件的且不形成环的..就加进生成树里..
View Code
1 #include <stdio.h> 2 #include <cstring> 3 #include <cmath> 4 #include <algorithm> 5 using namespace std; 6 #define clr(x) memset(x, 0, sizeof(x)) 7 int n, m, k; 8 9 int f[10510]; 10 int find(int x) 11 { 12 return f[x] == x?x:f[x] = find(f[x]); 13 } 14 15 struct Edge 16 { 17 int u, v; 18 char col; 19 }edge[1100005]; 20 21 int solve(bool flag) 22 { 23 int ans = 0; 24 char cc; 25 for(int i = 0; i <= n; ++i) 26 f[i] = i; 27 28 if(flag) 29 cc = 'B'; 30 else 31 cc = 'R'; 32 33 for(int i = 0; i < m; ++i) 34 if(edge[i].col == cc){ 35 int xx = find(edge[i].u), yy = find(edge[i].v); 36 if(xx != yy){ 37 f[xx] = yy; 38 ans++; 39 } 40 } 41 return ans; 42 } 43 44 int main() 45 { 46 int i, j; 47 int bl, re; 48 while(scanf("%d %d %d", &n, &m, &k) != EOF) 49 { 50 if(n == 0 && m == 0 && k == 0) break; 51 bl = re = 0; 52 53 for(i = 0; i < m; ++i){ 54 getchar(); 55 scanf("%c%d%d", &edge[i].col, &edge[i].u, &edge[i].v); 56 } 57 58 bl = solve(true); 59 re = solve(false); 60 if(k <= bl && k >= n-1-re) puts("1"); 61 else puts("0"); 62 } 63 return 0; 64 }