UVA 11865 Stream My Contest （最小树形图 + 二分）

题目链接：http://acm.hust.edu.cn/vjudge/problem/viewProblem.action?id=26164

题意：你需要花费不超过cost元来搭建一个比赛网络。网络中有n台机器，编号0~n-1，其中机器0为服务器，其他机器为客户机。一共有m条可以使用的网线，其中第i条网线的发送端是机器ui，接收端是机器vi（数据只能从机器ui单向传输到机器vi），带宽是bi Kbps,费用是ci元。每台客户机应当恰好从一台机器接收数据。你的任务是最大化网络中的最小带宽。

思路：二分带宽，看能否跑出最小树形图

#include <cstdlib>#include <cctype>#include <cstring>#include <cstdio>#include <cmath>#include <algorithm>#include <vector>#include <string>#include <iostream>#include <map>#include <set>#include <queue>#include <stack>#include <bitset>#include <functional>#include <utility>using namespace std;const int maxn = 100;const int maxm = 10010;const int inf = 0x3f3f3f3f;int pre[maxn], id[maxn], vis[maxn], in[maxn];int g[maxn][maxn], cnt, cnt1;int n, m, c;struct edge{    int u, v, w;    edge() {}    edge(int u, int v, int w) : u(u), v(v), w(w) {}} e[maxm];struct node{    int u, v, bit, w;} no[maxm];void add(int u, int v, int w){    e[cnt].u = u;    e[cnt].v = v;    e[cnt++].w = w;}void add1(int u, int v, int bit, int w){    no[cnt1].u = u;    no[cnt1].v = v;    no[cnt1].bit = bit;    no[cnt1++].w = w;}int zhuliu(int root, int n, int m){    int res = 0;    int u, v;    while(1)    {        for(int i = 0; i < n; i++)            in[i] = inf;        for(int i = 0; i < m; i++)            if(e[i].u != e[i].v && e[i].w < in[e[i].v])            {                pre[e[i].v] = e[i].u;                in[e[i].v] = e[i].w;            }        for(int i = 0; i < n; i++)            if(i != root && in[i] == inf)                return -1;        int tn = 0;        memset(id, -1, sizeof(id));        memset(vis, -1, sizeof(vis));        in[root] = 0;        for(int i = 0; i < n; i++)        {            res += in[i];            v = i;            while(vis[v] != i && id[v] == -1 && v != root)            {                vis[v] = i;                v = pre[v];            }            if(v != root && id[v] == -1)            {                for(int u = pre[v]; u != v; u = pre[u])                    id[u] = tn;                id[v] = tn++;            }        }        if(tn == 0) break;        for(int i = 0; i < n; i++)            if(id[i] == -1)                id[i] = tn++;        for(int i = 0; i < m;)        {            v = e[i].v;            e[i].u = id[e[i].u];            e[i].v = id[e[i].v];            if(e[i].u != e[i].v)                e[i++].w -= in[v];            else                swap(e[i], e[--m]);        }        n = tn;        root = id[root];    }    return res;}bool ok(int mid){    cnt = 0;    for(int i = 0; i < m; i++)        if(no[i].bit >= mid)        {            add(no[i].u, no[i].v, no[i].w);            //printf("%d %d %d\n", e[i].u, e[i].v, e[i].w);        }    int ans = zhuliu(0, n, cnt);    if(ans > c || ans == -1)        return false;    return true;}int main(){    int t;    scanf("%d", &t);    for(int ca = 1; ca <= t; ca++)    {        cnt = 0;        cnt1 = 0;        scanf("%d%d%d", &n, &m, &c);        memset(g, inf , sizeof(g));        for(int i = 0; i < m; i++)        {            int u, v, bit, w;            scanf("%d%d%d%d", &u, &v, &bit, &w);            add1(u, v, bit, w);            add(u, v, w);        }        int ans = zhuliu(0, n, cnt);        if(ans == -1)            puts("streaming not possible.");        else        {            int l = 0, r = 1e6, ans = 0;            while(l <= r)            {                int mid = (l + r) >> 1;                if(ok(mid))                    l = mid + 1, ans = mid;                else                    r = mid - 1;            }            if(!ans)                puts("streaming not possible.");            else                printf("%d kbps\n", ans);        }    }    return 0;}