CSU 1574 Amanda Lounges Simulation problem

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Test instructions

A graph of the n-point m edges (starting at each point is white)

The following M-line gives the edges and edges, and the edge weights indicate the number of dots that are dyed black in the 2 points connected by the edge.

0 means dye, 1 means one dianran, and 2 means dye.

Q: The minimum number of points to be dyed can meet the above edge rights. If there is no output impossible

The case of all edge weights 0 and 2 is processed first.

In this case, only the sub-graph with the edge weight of 1 is left.

Dye a point, and then the BFS to dye the map.

#include <iostream> #include <stdio.h> #include <string.h> #include <queue> #include <math.h > #include <vector>using namespace std;template <class t>inline BOOL Rd (T &ret) {char c; int sgn;if (C=ge TCHAR (), c==eof) return 0;while (c!= '-' && (c< ' 0 ' | | C> ' 9 ')) C=getchar (), sgn= (c== '-'), -1:1;ret= (c== '-')? 0: (c ' 0 '); while (C=getchar (), c>= ' 0 ' &&c<= ' 9 ') ret=ret*10+ (c ' 0 '); Ret*=sgn;return 1;}        Template <class t>inline void pt (T x) {if (x <0) {Putchar ('-');    x =-X;    } if (X>9) pt (X/10); Putchar (x%10+ ' 0 ');}  const int N = 200105;const int M = n<<1;typedef pair<int,int> pii;vector<pii>g;struct Edge{int from, to, Col, NEX;} edge[m];//Note N M to modify int head[n], edgenum;void Add (int u, int v, int col) {Edge E = {u, V, col, head[u]};edge[edgenum] = e;h Ead[u] = Edgenum + +;} void Init () {memset (head,-1, sizeof head); edgenum = 0;} int C[n], ans, N, m;int bfs (int x) {int siz = 1, ran = 1; c[x] = 1;queue<int>q;q.push (x);//printf ("bfs:%d ' s color =%d\n", x, C[x]), while (!q.empty ()) {int u = q.front (); Q.pop (); for (int i = Head[u ]; ~i; i = Edge[i].nex) {int v = edge[i].to;//printf ("Bfs_edge (%d,%d)-(%d,%d) \ n", u, C[u], V, C[v]), if (c[v]!=-1) {//printf ("sum = %d\n ", c[u]+c[v]); if ((C[u]+c[v))!=1) return-1;} else {Siz++;c[v] = C[u]^1;ran + = C[v];q.push (v);//printf ("%d s color =%d\n", V, C[v]);}} Return min (ran, Siz-ran);} BOOL Solve () {int u, V, D;memset (c,-1, sizeof C), for (int i = 0; i < m; i++) {rd (U); Rd (v); Rd (d); Add (U, V, d); add (V, U, d);} queue<int>q;for (int i = 0; i < Edgenum; i+=2) {u = edge[i].from; v = edge[i].to; d = edge[i].col;if (d = = 0) {if (c[u ]==1 | | C[V] = = 1) return false;if (c[u]==-1) Q.push (U), c[u] = 0;if (c[v]==-1) Q.push (v), c[v] = 0;} else if (d = = 2) {if (c[u] = = 0 | | c[v] = = 0) return false;if (c[u]==-1) Q.push (U), c[u] = 1;if (c[v]==-1) Q.push (v), c[v] = 1;}} while (!q.empty ()) {u = Q.front (), Q.pop (), for (int i = Head[u];~i;i=edge[i].nex) {v = edge[i].to;if (c[v]!=-1) {if (Edge[i]. Col! = C[u]+c[v]) return false;} else {C[v] = C[u]^1;q.push (v);}}} Puts ("init:"); for (int i = 1; I <= N; i++) printf ("(%d ' color is%d) \ n", I, C[i]); for (int i = 1; I <= n; i++) ans + = c[i] = = 1; G.clear (); for (int i = 0; i < Edgenum; i+=2) {if (Edge[i].col! = 1) continue;u = edge[i].from; v = edge[i].to;if (C[u]==-1 & amp;& c[v]==-1) G.push_back (PII (U,v));}  Init (); for (int i = 0; i < (int) g.size (); i++) {u = g[i].first; v = g[i].second; add (u,v,1); add (v,u,1);//printf ("Addedge (%d,%d) \ n ", u, v);} for (int i = 1; I <= n; i++) if (c[i] = = 1) {int tmp = BFS (i); if (tmp = =-1) return False;ans + = tmp; return true;} int main () {while (~SCANF ("%d%d", &n, &m)) {init (); ans = 0;if (false = = Solve ()) puts ("impossible"); else cout< <ans<<endl;} return 0;} /*5 2 (3) 4 5 1ans:35 2 3 4 5 5 2 1ANS:IM5 3 4 5 6 6 81 2 22 0ans:39 3 13 4 5 6 to 6 8 9 1ans:49 2 3 4 5 6 6 8 9 1ans:im*/

CSU 1574 Amanda Lounges Simulation problem