CSU 1574 Amanda Lounges 模拟题

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题目链接：点击打开链接

题意：

给定n个点m条边的无向图（开始每个点都是白色）

下面m行给出边和边权，边权表示这条边所连接的2个点中被染成黑色的点数。

0表示染，1表示其中一个点染，2表示都染。

问：最少染多少个点可以满足上述的边权。若不存在输出impossible

首先处理所有边权为0和2的情况。

这样处理后图中就只剩下边权为1的子图。

任意染一个点，然后bfs一下把子图染掉即可。

#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=getchar(),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];//注意 N M 要修改int head[N], edgenum;void add(int u, int v, int col){Edge E = {u, v, col, head[u]};edge[edgenum] = E;head[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 && 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 41 2 22 3 13 4 14 5 1ans:35 51 2 22 3 13 4 14 5 13 5 1ans:im5 61 2 22 3 13 4 14 5 13 6 05 6 0ans:39 81 2 22 3 13 4 14 5 13 6 05 6 07 8 18 9 1ans:49 91 2 22 3 13 4 14 5 13 6 05 6 07 8 18 9 19 7 1ans:im*/

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