Tricks Device (hdu 5294 最短路+最大流)
Tricks Device (hdu 5294 最短路+最大流)
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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 1Sample Output
2 6Source
We have carefully selected several similar problems for you: 5299 5296
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;
}发布者：全栈程序员栈长，转载请注明出处：https://javaforall.cn/115451.html原文链接：https://javaforall.cn