HDU 2121 Ice_cream’s world II

Ice_cream’s world II

Time Limit: 1000ms

Memory Limit: 32768KB

This problem will be judged on HDU. Original ID: 2121
64-bit integer IO format: %I64d      Java class name: Main

 

After awarded lands to ACMers, the queen want to choose a city be her capital. This is an important event in ice_cream world, and it also a very difficult problem, because the world have N cities and M roads, every road was directed. Wiskey is a chief engineer in ice_cream world. The queen asked Wiskey must find a suitable location to establish the capital, beautify the roads which let capital can visit each city and the project’s cost as less as better. If Wiskey can’t fulfill the queen’s require, he will be punishing.

 

Input

Every case have two integers N and M (N<=1000, M<=10000), the cities numbered 0…N-1, following M lines, each line contain three integers S, T and C, meaning from S to T have a road will cost C.

 

Output

If no location satisfy the queen’s require, you must be output “impossible”, otherwise, print the minimum cost in this project and suitable city’s number. May be exist many suitable cities, choose the minimum number city. After every case print one blank.

 

Sample Input

3 1
0 1 1

4 4
0 1 10
0 2 10
1 3 20
2 3 30

Sample Output

impossible

40 0

Source

HDU 2007-10 Programming Contest_WarmUp

 

解题：最小树形图。由于我们加的虚拟根与边有关，所以可以方便的找出实际根

 

#include <bits/stdc++.h>
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 1010;
struct arc {
    int u,v,w;
} e[50000];
int in[maxn],pre[maxn],hs[maxn],vis[maxn],Rt;
int DMST(int root,int n,int m,int ret = 0) {
    while(true) {
        for(int i = 0; i < n; ++i) {
            hs[i] = vis[i] = -1;
            in[i] = INF;
        }
        for(int i = 0; i < m; ++i) {
            if(e[i].u != e[i].v && e[i].w < in[e[i].v]) {
                in[e[i].v] = e[i].w;
                pre[e[i].v] = e[i].u;
                if(e[i].u == root) Rt = i;
            }
        }
        for(int i = 0; i < n; ++i)
            if(root != i && in[i] == INF) return -1;
        int cnt = in[root] = 0;
        for(int i = 0; i < n; ++i) {
            ret += in[i];
            int v = i;
            while(vis[v] != i && hs[v] == -1 && v != root) {
                vis[v] = i;
                v = pre[v];
            }
            if(v != root && hs[v] == -1) {
                for(int u = pre[v]; u != v; u = pre[u]) hs[u] = cnt;
                hs[v] = cnt++;
            }
        }
        if(!cnt) break;
        for(int i = 0; i < n; ++i)
            if(hs[i] == -1) hs[i] = cnt++;
        for(int i = 0; i < m; ++i) {
            int u = e[i].u;
            int v = e[i].v;
            e[i].u = hs[u];
            e[i].v = hs[v];
            if(e[i].u != e[i].v) e[i].w -= in[v];
        }
        n = cnt;
        root = hs[root];
    }
    return ret;
}
int main() {
    int n,m,sum;
    while(~scanf("%d%d",&n,&m)) {
        for(int i = sum = 0; i < m; ++i) {
            scanf("%d%d%d",&e[i].u,&e[i].v,&e[i].w);
            sum += e[i].w;
        }
        sum++;
        for(int i = 0; i < n; ++i) {
            e[i + m].u = n;
            e[i + m].v = i;
            e[i + m].w = sum;
        }
        int ret = DMST(n,n + 1,n + m);
        if(ret == -1 || ret > (sum<<1)) puts("impossible");
        else printf("%d %d\n",ret-sum,Rt - m);
        puts("");
    }
    return 0;
}