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解题报告

http://blog.csdn.net/juncoder/article/details/38349019

题目传送门

题意：

思路：

二分容量建图，spfa判时间是否符合条件

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define inf 99999999
using namespace std;
int n,m,T;
struct node {
    int u,v,w,t,next;
} edge[501000],N[101000];
int head[101000],cnt;
void add(int u,int v,int w)
{
    edge[cnt].v=v;
    edge[cnt].t=w;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
int spfa()
{
    int i,j,dis[10100],vis[10100];
    for(i=1; i<=n; i++) {
        dis[i]=inf;
        vis[i]=0;
    }
    dis[1]=0;
    vis[1]=1;
    queue<int>Q;
    Q.push(1);
    while(!Q.empty()) {
        int u=Q.front();
        Q.pop();
        vis[u]=0;
        for(i=head[u]; i!=-1; i=edge[i].next) {
            if(dis[edge[i].v]>dis[u]+edge[i].t) {
                dis[edge[i].v]=dis[u]+edge[i].t;
                if(!vis[edge[i].v]) {
                    vis[edge[i].v]=1;
                    Q.push(edge[i].v);
                }
            }
        }
    }
    if(dis[n]<=T)
        return 1;
    return 0;
}
int slove(int mid)
{
    memset(edge,0,sizeof(edge));
    memset(head,-1,sizeof(head));
    cnt=0;
    for(int i=1; i<=m; i++) {
        if(N[i].w>=mid) {
            add(N[i].u,N[i].v,N[i].t);
            add(N[i].v,N[i].u,N[i].t);
        }
    }
    if(spfa())return 1;
    return 0;
}
int main()
{
    int tt,i,j;
    scanf("%d",&tt);
    while(tt--) {
        memset(N,0,sizeof(N));
        scanf("%d%d%d",&n,&m,&T);
        int l=0,r=-1;
        for(i=1; i<=m; i++) {
            scanf("%d%d%d%d",&N[i].u,&N[i].v,&N[i].w,&N[i].t);
            if(N[i].w>r)
                r=N[i].w;
        }
        int maxx=-1;
        while(l<=r) {
            int mid=(l+r)/2;
            if(slove(mid)) {
                l=mid+1;
                maxx=mid;
            } else r=mid-1;
        }
        printf("%d\n",maxx);
    }
    return 0;
}

Delay Constrained Maximum Capacity Path

2
2 1 10
1 2 13 10
4 4 20
1 2 1000 15
2 4 999 6
1 3 100 15
3 4 99 4

13
99

HDU1839_Delay Constrained Maximum Capacity Path(最短路+二分),布布扣,bubuko.com

HDU1839_Delay Constrained Maximum Capacity Path(最短路+二分)

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原文地址：http://blog.csdn.net/juncoder/article/details/38349019