Tricks Device

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 124    Accepted Submission(s): 27

Problem Description
Innocent Wu follows Dumb Zhang into a ancient tomb. Innocent Wu’s at the entrance of the tomb while Dumb Zhang’s at the end of it. The tomb is made up of many chambers, the total number is N. And there are M channels connecting the
chambers. Innocent Wu wants to catch up Dumb Zhang to find out the answers of some questions, however, it’s Dumb Zhang’s intention to keep Innocent Wu in the dark, to do which he has to stop Innocent Wu from getting him. Only via the original shortest ways
from the entrance to the end of the tomb costs the minimum time, and that’s the only chance Innocent Wu can catch Dumb Zhang.

Unfortunately, Dumb Zhang masters the art of becoming invisible(奇门遁甲) and tricks devices of this tomb, he can cut off the connections between chambers by using them. Dumb Zhang wanders how many channels at least he has to cut to stop Innocent Wu. And Innocent
Wu wants to know after how many channels at most Dumb Zhang cut off Innocent Wu still has the chance to catch Dumb Zhang.
 
Input
There are multiple test cases. Please process till EOF.

For each case,the first line must includes two integers, N(<=2000), M(<=60000). N is the total number of the chambers, M is the total number of the channels.

In the following M lines, every line must includes three numbers, and use ai、bi、li as channel i connecting chamber ai and bi(1<=ai,bi<=n), it costs li(0<li<=100) minute to pass channel i.

The entrance of the tomb is at the chamber one, the end of tomb is at the chamber N.
 
Output
Output two numbers to stand for the answers of Dumb Zhang and Innocent Wu’s questions.
 
Sample Input
8 9
1 2 2
2 3 2
2 4 1
3 5 3
4 5 4
5 8 1
1 6 2
6 7 5
7 8 1
 
Sample Output
2 6
 
Source
 
Recommend
We have carefully selected several similar problems for you:  5299 

pid=5298" target="_blank">5298 

pid=5297" target="_blank">5297 5296 

pid=5295" target="_blank">5295 

 

题意:n个点m条无向边,如果从起点0到终点n-1的最短路距离为dist,求最少删除多少条边使得图中不再存在最短路。最多删除多少条边使得图中仍然存在最短路。

思路:先用spfa求一次最短路,开一个road数组,road[i]表示从起点走到i点最短路径所经过的最少边数,然后第二问就是m-road[n-1];再依据最短路的dist数组推断哪些边是最短路上的,用它们又一次构图。跑一遍网络流求最小割。比赛的时候没有在最短路上建边,直接用的原图。果断TLE,又坑了队友=-=

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string>
#include <map>
#include <stack>
#include <vector>
#include <set>
#include <queue>
#pragma comment (linker,"/STACK:102400000,102400000")
#define pi acos(-1.0)
#define eps 1e-6
#define lson rt<<1,l,mid
#define rson rt<<1|1,mid+1,r
#define FRE(i,a,b) for(i = a; i <= b; i++)
#define FREE(i,a,b) for(i = a; i >= b; i--)
#define FRL(i,a,b) for(i = a; i < b; i++)
#define FRLL(i,a,b) for(i = a; i > b; i--)
#define mem(t, v) memset ((t) , v, sizeof(t))
#define sf(n) scanf("%d", &n)
#define sff(a,b) scanf("%d %d", &a, &b)
#define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c)
#define pf printf
#define DBG pf("Hi\n")
typedef long long ll;
using namespace std; #define INF 0x3f3f3f3f
#define mod 1000000009
const int MAXN = 2005;
const int MAXM = 200005;
const int N = 1005; int n,m; struct EDGE
{
int u,v,len,next;
}e[MAXM]; struct Edge
{
int to,next,cap,flow;
}edge[MAXM]; int tol;
int head[MAXN]; void init()
{
tol=0;
memset(head,-1,sizeof(head));
} void add(int u,int v,int len)
{
e[tol].u=u;
e[tol].v=v;
e[tol].len=len;
e[tol].next=head[u];
head[u]=tol++;
e[tol].u=v;
e[tol].v=u;
e[tol].len=len;
e[tol].next=head[v];
head[v]=tol++;
} void addedge(int u,int v,int w,int rw=0)
{
edge[tol].to=v;
edge[tol].cap=w;
edge[tol].flow=0;
edge[tol].next=head[u];
head[u]=tol++; edge[tol].to=u;
edge[tol].cap=rw;
edge[tol].flow=0;
edge[tol].next=head[v];
head[v]=tol++;
} int Q[MAXN];
int dep[MAXN],cur[MAXN],sta[MAXN]; bool bfs(int s,int t,int n)
{
int front=0,tail=0;
memset(dep,-1,sizeof(dep[0])*(n+1));
dep[s]=0;
Q[tail++]=s;
while (front<tail)
{
int u=Q[front++];
for (int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].to;
if (edge[i].cap>edge[i].flow && dep[v]==-1)
{
dep[v]=dep[u]+1;
if (v==t) return true;
Q[tail++]=v;
}
}
}
return false;
} int dinic(int s,int t,int n)
{
int maxflow=0;
while (bfs(s,t,n))
{
for (int i=0;i<n;i++) cur[i]=head[i];
int u=s,tail=0;
while (cur[s]!=-1)
{
if (u==t)
{
int tp=INF;
for (int i=tail-1;i>=0;i--)
tp=min(tp,edge[sta[i]].cap-edge[sta[i]].flow);
maxflow+=tp;
for (int i=tail-1;i>=0;i--)
{
edge[sta[i]].flow+=tp;
edge[sta[i]^1].flow-=tp;
if (edge[sta[i]].cap-edge[sta[i]].flow==0)
tail=i;
}
u=edge[sta[tail]^1].to;
}
else if (cur[u]!=-1 && edge[cur[u]].cap > edge[cur[u]].flow &&dep[u]+1==dep[edge[cur[u]].to])
{
sta[tail++]=cur[u];
u=edge[cur[u]].to;
}
else
{
while (u!=s && cur[u]==-1)
u=edge[sta[--tail]^1].to;
cur[u]=edge[cur[u]].next;
}
}
}
return maxflow;
} int dist[MAXN];
int vis[MAXN];
int road[MAXN]; void SPFA()
{
memset(vis,0,sizeof(vis));
memset(dist,INF,sizeof(dist));
memset(road,INF,sizeof(road));
dist[0]=0;
road[0]=0;
vis[0]=1;
queue<int>Q;
Q.push(0);
while (!Q.empty())
{
int u=Q.front();
Q.pop();
vis[u]=0;
for (int i=head[u];~i;i=e[i].next)
{
int v=e[i].v;
if (dist[v]>dist[u]+e[i].len)
{
dist[v]=dist[u]+e[i].len;
road[v]=road[u]+1;
if (!vis[v])
{
vis[v]=1;
Q.push(v);
}
}
else if (dist[v]==dist[u]+e[i].len)
{
if (road[v]>road[u]+1)
{
road[v]=road[u]+1;
if (!vis[v])
{
vis[v]=1;
Q.push(v);
}
}
}
}
}
} int main()
{
#ifndef ONLINE_JUDGE
freopen("C:/Users/lyf/Desktop/IN.txt","r",stdin);
#endif
int i,j,u,v,w;
while (~sff(n,m))
{
init();
for (i=0;i<m;i++)
{
sfff(u,v,w);
if (u==v) continue;
u--;v--;
add(u,v,w);
}
SPFA();
int cnt=tol;
init();
for (i=0;i<cnt;i++)
{
u=e[i].u;
v=e[i].v;
if (dist[v]==dist[u]+e[i].len)
addedge(u,v,1);
}
int ans=dinic(0,n-1,n);
pf("%d %d\n",ans,m-road[n-1]);
}
return 0;
}

Tricks Device (hdu 5294 最短路+最大流)的更多相关文章

  1. hdu 5294 最短路+最大流 ***

    处理处最短路径图,这个比较巧妙 链接:点我

  2. HDU 5294 Tricks Device(多校2015 最大流+最短路啊)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5294 Problem Description Innocent Wu follows Dumb Zha ...

  3. hdu 3599(最短路+最大流)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3599 思路:首先spfa求一下最短路,然后对于满足最短路上的边(dist[v]==dist[u]+w) ...

  4. HDU 5294 Tricks Device (最大流+最短路)

    题目链接:HDU 5294 Tricks Device 题意:n个点,m条边.而且一个人从1走到n仅仅会走1到n的最短路径.问至少破坏几条边使原图的最短路不存在.最多破坏几条边使原图的最短路劲仍存在 ...

  5. hdu 5294 Tricks Device 最短路建图+最小割

    链接:http://acm.hdu.edu.cn/showproblem.php?pid=5294 Tricks Device Time Limit: 2000/1000 MS (Java/Other ...

  6. HDU 5294 Tricks Device 网络流 最短路

    Tricks Device 题目连接: http://acm.hdu.edu.cn/showproblem.php?pid=5294 Description Innocent Wu follows D ...

  7. HDOJ 5294 Tricks Device 最短路(记录路径)+最小割

    最短路记录路径,同一时候求出最短的路径上最少要有多少条边, 然后用在最短路上的边又一次构图后求最小割. Tricks Device Time Limit: 2000/1000 MS (Java/Oth ...

  8. SPFA+Dinic HDOJ 5294 Tricks Device

    题目传送门 /* 题意:一无向图,问至少要割掉几条边破坏最短路,问最多能割掉几条边还能保持最短路 SPFA+Dinic:SPFA求最短路时,用cnt[i]记录到i最少要几条边,第二个答案是m - cn ...

  9. HDU5294 Tricks Device(最大流+SPFA) 2015 Multi-University Training Contest 1

    Tricks Device Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) To ...

随机推荐

  1. 一种基于Storm的可扩展即时数据处理架构思考

    问题引入 使用storm可以方便的构建一种集群式的数据框架,并通过定义topo来实现业务逻辑. 但使用topo存在一个缺点, topo的处理能力来自于其启动时设置的worker数目,在很多情况下,我们 ...

  2. bzoj1084: [SCOI2005]最大子矩阵

    dp.状态转移方程在代码里 #include<cstdio> #include<algorithm> #include<cstring> using namespa ...

  3. [原]Unity3D深入浅出 - 物理材质(Physics Materials)

    在Unity3d中已经配置好了5种常用的物理材质,Bouncy.Ice.Metal.Rubber.Wood,在菜单中依次选择Assets - Import Package - Physics Mate ...

  4. 嵌入式linux市场份额

    来自华清远见2014年度的调查统计数据显示,在嵌入式产品研发的软件开发平台的选择上,嵌入式Linux以55%的市场份额遥遥领先于其他嵌入式开发软件发平台,比去年增长了13个百分比,这已经是连续4年比例 ...

  5. Kooboo CMS的安装步骤

    Kooboo CMS的安装步骤 来自Kooboo document 跳转到: 导航, 搜索 http://www.microsoft.com/web/gallery/install.aspx?appi ...

  6. HDU 5965 Gym Class 贪心+toposort

    分析:就是给一些拓补关系,然后求最大分数,所以贪心,大的越靠前越好,小的越靠后越好 剩下的就是toposort,当然由于贪心,所以使用优先队列 #include <iostream> #i ...

  7. HDU 3533 Escape BFS搜索

    题意:懒得说了 分析:开个no[100][100][1000]的bool类型的数组就行了,没啥可说的 #include <iostream> #include <cstdio> ...

  8. Web API-如何将Controller的返回值转换成HTTP response消息

    https://www.asp.net/web-api/overview/formats-and-model-binding/json-and-xml-serialization https://co ...

  9. Git管理命令

     1.创建分支 git branch <分支名> 2.切换分支 git checkout <分支名> 创建并切换到该分支:git checkout -b <分支名> ...

  10. python错误收集

    Installing 'flask'You are using pip version 6.1.1, however version 7.1.2 is available.You should con ...