hdu4263 Red/Blue Spanning Tree

思路：有两种没有边权但是有染色的边，现在问时候能够形成确切的含有K条Blue边的生成树。直接找是不现实的，但是我们可以看至少和之多需要多少条就行了，只要K在这两个值之间就行了。因为从最少的Blue边开始，用blue边替换RED边，这样是可以形成环的，所以替换是可行的，这样就可以一步一步替换成含有k条blue边的生成树。

// #pragma comment(linker, "/STACK:1024000000,1024000000")#include <iostream>#include <algorithm>#include <iomanip>#include <sstream>#include <string>#include <stack>#include <queue>#include <deque>#include <vector>#include <map>#include <set>#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <climits>using namespace std;// #define DEBUG#ifdef DEBUG#define debug(...) printf( __VA_ARGS__ )#else#define debug(...)#endif#define CLR(x) memset(x, 0,sizeof x)#define MEM(x,y) memset(x, y,sizeof x)#define pk push_backtemplate<class T> inline T Get_Max(const T&a,const T&b){return a < b?b:a;}template<class T> inline T Get_Min(const T&a,const T&b){return a < b?a:b;}typedef long long LL;typedef unsigned long long ULL;typedef pair<int,int> ii;const double eps = 1e-10;const int inf = 1 << 30;const int INF = 0x3f3f3f3f;const int MOD = 1e9 + 7;const int maxn = 1010;int n,m,k;struct Edge{int u, v, w;Edge(){}Edge(int _u,int _v,int _w):u(_u),v(_v),w(_w){}bool operator < (const Edge& rhs)const{return w < rhs.w;}}E[maxn*10000];int F[maxn];int Find(int x){if (F[x]==-1) return x;return F[x] = Find(F[x]);}int cnt;int solve(){memset(F, -1,sizeof F);int ans = 0;int sum = 0;sort(E,E+cnt);for (int i = 0;i < cnt;++i){int t1 = Find(E[i].u);int t2 = Find(E[i].v);if (t1 != t2){F[t1] = t2;sum += E[i].w;ans++;}if (ans==n-1)return sum;}if (ans==n-1) return sum;return INF;}int main(){// freopen("in.txt","r",stdin);// freopen("out.txt","w",stdout);while(scanf("%d%d%d",&n,&m,&k) != EOF){if (n==0&&m==0&&k==0)break;cnt=0;char op[10];int x,y;for (int i = 0;i < m;++i){scanf("%s",op);scanf("%d%d",&x,&y);E[cnt++] = Edge(x,y,op[0]=='B'?0:1);}int ans = solve();for (int i = 0;i < cnt;++i){E[i].w = 1 - E[i].w;}int sum = solve();if (ans != INF){ans = (n-1) - ans;if (sum <= k && k <= ans){cout << 1 << endl;continue;}}cout << 0 << endl;}return 0;}