hdu 1839 Delay Constrained Maximum Capacity Path 二分+ spfa

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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[N], dis[N], vis[N];int cap[N];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[n] == inf || dis[n] > t )                l = mid + 1;            else                r = mid;        }        printf("%d\n", cap[l]);    }    return 0;}

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