Uva 11865 - Stream My Contest（二分+朱刘算法）

最新推荐文章于 2018-09-15 10:37:33 发布

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最新推荐文章于 2018-09-15 10:37:33 发布

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分类专栏：二分图论-----------生成树文章标签：二分朱刘算法

本文链接：https://blog.youkuaiyun.com/xiao_k666/article/details/79548694

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本文探讨了一道经典的图论问题：在有限预算下寻找有向图中从根节点到其他节点的路径，使得路径总长度最大且不超过指定成本。采用二分查找结合朱刘算法实现解决方案。

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题目链接 https://vjudge.net/problem/UVA-11865

【题意】
有一张n个顶点，m条边的有向图，根节点为0。每条边有两个权值，一个是费用c，一个是长度b。问在总费用不超过cost的情况下选出若干条边，使得n个点连通时的边的最短长度的最大值是多少。

【思路】
二分+朱刘算法，对最小带宽进行二分，看看能否只用大于等于最小带宽的网线连接所有机器同时不会超出预算费用

#include<bits/stdc++.h>
using namespace std;

const int inf = 2e9;
const int maxn = 100;
const int maxm = 10050;

int n, m, cost;
int le, ri, mid;
int in[maxn], pre[maxn];
int used[maxn], id[maxn];

struct Edge {
    int from, to, b, c;
    bool ok = true;
    Edge(int f = 0, int t = 0, int bb = 0, int cc = 0) :from(f), to(t), b(bb), c(cc) {}
}edges[maxm], cpy[maxm];

int direct_mst(int root, int V, int E) {
    int ans = 0;
    while (1) { 
        for (int i = 0; i < V; ++i) in[i] = inf;
        for (int i = 0; i < E; ++i) {
            if (!edges[i].ok) continue;
            int u = edges[i].from;
            int v = edges[i].to;
            if (in[v] > edges[i].c && u != v) {
                in[v] = edges[i].c;
                pre[v] = u;
            }
        }

        for (int i = 0; i < V; ++i) {
            if (i == root) continue;
            if (inf == in[i]) return -1;
        }

        int cnt = 0;
        memset(id, -1, sizeof(id)); 
        memset(used, -1, sizeof(used));
        in[root] = 0;
        for (int i = 0; i < V; ++i) {
            ans += in[i];
            int v = i;
            while (used[v] != i && id[v] == -1 && v != root) {
                used[v] = i;
                v = pre[v];
            }
            if (v != root && id[v] == -1) {
                for (int u = pre[v]; u != v; u = pre[u]) id[u] = cnt;
                id[v] = cnt++;
            }
        }

        if (0 == cnt) break;

        for (int i = 0; i < V; i++)
            if (id[i] == -1) id[i] = cnt++;

        for (int i = 0; i < E; i++) {
            if (!edges[i].ok) continue;
            int u = edges[i].from;
            int v = edges[i].to;
            edges[i].from = id[u];
            edges[i].to = id[v];
            if (id[u] != id[v]) edges[i].c -= in[v];
        }
        V = cnt;
        root = id[root];
    }
    return ans;
}

bool check(int x) {
    memcpy(edges, cpy, sizeof(edges));
    for (int i = 0; i < m; ++i) {
        if (edges[i].b >= x) edges[i].ok = true;
        else edges[i].ok = false;
    }
    int ans = direct_mst(0, n, m);
    return ans != -1 && ans <= cost;
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d%d", &n, &m, &cost);
        le = inf, ri = 0;
        for (int i = 0; i < m; ++i) {
            int u, v, b, c;
            scanf("%d%d%d%d", &u, &v, &b, &c);
            edges[i] = Edge(u, v, b, c);
            cpy[i] = edges[i];
            le = min(le, b);
            ri = max(ri, b);
        }
        if (!check(le)) {
            printf("streaming not possible.\n");
            continue;
        }
        while (le + 1 < ri) {
            mid = (le + ri) >> 1;
            if (check(mid)) le = mid;
            else ri = mid;
        }
        printf("%d kbps\n", le);
    }
    return 0;
}