hdu 1839 Delay Constrained Maximum Capacity Path 二分+ spfa

有N个点，点1为珍贵矿物的采矿区， 点N为加工厂，有M条双向连通的边连接这些点。走每条边的运输容量为C，运送时间为D。选择一条从1到N的路径运输， 这条路径的运输总时间要在T之内，在这个前提之下，要让这条路径的运输容量尽可能地大。一条路径的运输容量取决与这条路径中的运输容量最小的那条边。

带限制的最短路---

由于每条路都是有一定容量的，相对来说，容量越大那么满足的边就越少。就可以二分容量来求出满足的最大容量了。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<queue>

using namespace std;

const int N = 100005;
const int inf = 1 << 28;

struct node{
    int to, nxt, cost, cap;
}e[N*5];

int head
, dis
, vis
;
int cap
;
int cnt;
int n, m, t;
int limt;
void init()
{
    cnt = 0;
    memset(head, -1, sizeof(head));
}

bool cmp(int a, int b)
{
    return a > b;
}

void add( int u, int v, int cap, int cost )
{
    e[cnt].to = v;
    e[cnt].cap = cap;
    e[cnt].cost = cost;
    e[cnt].nxt = head[u];
    head[u] = cnt++;
    e[cnt].to = u;
    e[cnt].cap = cap;
    e[cnt].cost = cost;
    e[cnt].nxt = head[v];
    head[v] = cnt++;
}

void spfa()
{
    queue<int> q;
    while(!q.empty())
        q.pop();
    memset(vis, 0, sizeof(vis));
    for( int i = 1; i <= n; i++ )
        dis[i] = inf;
    dis[1] = 0;
    vis[1] = 1;
    q.push(1);
    while( !q.empty() )
    {
        int now = q.front();
        q.pop();
        vis[now] = 0;
        for( int i = head[now]; ~i; i = e[i].nxt )
        if(e[i].cap >= limt)
        {
            int to = e[i].to;
            if( dis[to] > dis[now] + e[i].cost )
            {
                dis[to] = dis[now] + e[i].cost;
                if( !vis[to] )
                {
                    vis[to] = 1;
                    q.push(to);
                }
            }
        }
    }
}

int main()
{
    int tot;
    for( scanf("%d", &tot); tot--; )
    {
        scanf("%d%d%d", &n, &m, &t);
        init();
        int u, v, cp, tt;
        for( int i = 1; i <= m; i++ )
        {
            scanf("%d%d%d%d", &u, &v, &cp, &tt);
            cap[i] = cp;
            add(u, v, cp, tt);
        }
        sort(cap+1, cap+1+m, cmp);
        int l = 1, r = m;
        while(l < r)
        {
            int mid = (l + r) >> 1;
            limt = cap[mid];
            spfa();
            if( dis
 == inf || dis
 > t )
                l = mid + 1;
            else
                r = mid;
        }
        printf("%d\n", cap[l]);
    }
    return 0;
}