hdu 2121

来源：互联网发布：js收藏地址编辑：程序博客网时间：2024/06/05 00:22

题意：有向图的最小生成树（只管出去，不管回来）标准的模板啊，
设置一个虚拟节点n 权值是max+1; 代码是对这别人的打的，第一次不会写！

#include <iostream>#include <cstdio>#include <cstring>using namespace std;#define N 1010#define inf 101000000struct edg {    int u,v,val;}data[N*N];int ansi; int in[N];int visit[N],id[N];int pre[N];int dri_mst(int root,int n,int m) {    int cnt = 0;    while(1) {        int u,v;        for(int i = 0; i < n; i++) in[i] = inf;        for(int i = 0; i < m; i++)  {   //in[i] 选择进入边权值最小的            u = data[i].u;            v = data[i].v;            if(data[i].val < in[v] && u != v) {                pre[v] = u;                if(u == root) ansi = i;                in[v] = data[i].val;            }        }        for(int i = 0; i < n; i++) {            if(i == root) continue;  // 根节点没有入度            if(in[i] == inf) return -1;  //说明存在点没有入边        }        int cntcode = 0; //环的个数        memset(visit,-1,sizeof(visit));        memset(id,-1,sizeof(id));        in[root] = 0;        for(int i = 0; i < n; i++) {            cnt += in[i];            v = i;            while(visit[v] != i && id[v] == -1&& v != root) {                visit[v] = i;                v = pre[v];            }            if(v != root && id[v] == -1) {                for(u = pre[v]; u != v; u = pre[u])                   id[u] = cntcode;  // u属于哪个环                id[v] = cntcode++;            }        }        if(cntcode == 0) break;  // 不存在环  //生成树构成了        for(int i = 0; i < n; i++) // 缩点 重新标记          if(id[i] == -1)            id[i] = cntcode++;        for(int i = 0; i < m; i++) {            v = data[i].v;            data[i].u = id[data[i].u];            data[i].v = id[data[i].v];            if(data[i].u != data[i].v)  // 自环              data[i].val -= in[v];        }        n = cntcode;         root = id[root];    }    return cnt;}int main() {    int n,m;    while(~scanf("%d %d",&n,&m)) {        int sum = 0;        for(int i = 0; i < m; i++) {            scanf("%d %d %d",&data[i].u,&data[i].v,&data[i].val);            sum += data[i].val;        }        sum ++;        for(int i = m; i < m + n; i++) {            data[i].u = n;            data[i].v = i - m;            data[i].val = sum;        }        int ans = dri_mst(n,n+1,n+m);        if(ans == -1 || ans >= 2*sum) printf("impossible\n\n");        else printf("%d %d\n\n",ans - sum,ansi-m);    }    return 0;}

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