（寒假集训）Roadblock（最短路）

Roadblock

时间限制: 1 Sec 内存限制: 64 MB
提交: 9 解决: 5
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题目描述

Every
morning, FJ wakes up and walks across the farm from his house to the
barn. The farm is a collection of N fields (1 <= N <= 250)
connected by M bidirectional pathways (1 <= M <= 25,000), each
with an associated length.
FJ's house is in field 1, and the barn is
in field N. No pair of fields is joined by multiple redundant pathways,
and it is possible to travel between any pair of fields in the farm by
walking along an appropriate sequence of pathways.
When traveling from one field to another, FJ always selects a route
consisting of a sequence of pathways having minimum total length.

Farmer John's cows, up to no good as
always, have decided to interfere with his morning routine. They plan
to build a pile of hay bales on exactly one of the M pathways on the
farm, doubling its length. The cows wish to select
a pathway to block so that they maximize the increase in FJ's distance
from the house to the barn. Please help the cows determine by how much
they can lengthen FJ's route.

输入

* Line 1: Two space-separated integers, N and M.
* Lines 2..1+M: Line j+1 describes the jth
bidirectional pathway in terms of three space-separated integers: A_j
B_j L_j, where A_j and B_j are indices in the range 1..N indicating
the fields joined by the pathway, and L_j is the length of the pathway
(in the range 1...1,000,000).

输出

*
Line 1: The maximum possible increase in the total length of FJ's
shortest route made possible by doubling the length of a single
pathway.

样例输入

样例输出

提示

There are 5 fields and 7 pathways. Currently, the shortest path from the house (field 1) to the barn (field 5) is 1-3-4-5 of total length 1+3+2=6.If the cows double the length of the pathway from field 3 to field 4 (increasing its length from 3 to 6), then FJ's shortest route is now 1-3-5, of total length 1+7=8, larger by two than the previous shortest route length.

【分析】农夫要从1走到n，他只走最短路。现在他的牛想使坏，想在一些路上设置障碍，使得这条路的长度变为2倍，求最短路的最大增量（农夫选择的）。先跑一边最段路，找前驱，然后将前驱所练成的每一条边依次遍历将其扩大二倍，再在只把该边扩大二倍的原图上跑最短路，找d[n]的最大值。可能存在很多的最短路，但是只用处理一条，因为即使有两条不相交的最短路，那么跑完之后d[n]还跟第一次跑的一样，不影响。

#include <iostream>

#include <cstring>

#include <cstdio>

#include <algorithm>

#include <cmath>

#include <string>

#include <map>

#include <stack>

#include <queue>

#include <vector>

#define inf 0x3f3f3f3f

#define met(a,b) memset(a,b,sizeof a)

#define pb push_back

typedef long long ll;

using namespace std;

const int N = 1e3;

const int M = ;

int n,m,k;

int edgg[N][N];

int edg[N][N],vis[N],d[N],pre[N];

void spfa(int x)

{

    met(vis,);

    met(d,inf);

    d[]=;

    vis[]=;

    queue<int>q;

    q.push();

    while(!q.empty()){

        int t=q.front();q.pop();

        vis[t]=;

        for(int i=;i<=n+;i++){

            if(d[i]>d[t]+edg[t][i]){

                d[i]=d[t]+edg[t][i];

                if(!x)pre[i]=t;

                if(!vis[i])q.push(i),vis[i]=;

            }

        }

    }

}

int main()

{

    met(edg,inf);

    met(pre,-);

    int ans1,ans2=;

    scanf("%d%d",&n,&m);

    int u,v,w;

    while(m--){

        scanf("%d%d%d",&u,&v,&w);

        edg[u][v]=edg[v][u]=w;

    }

    spfa();

    ans1=d[n];

    for(int i=n;pre[i]!=-;i--){

        int v=pre[i];

        edg[i][v]*=;

        edg[v][i]*=;

        spfa();

        ans2=max(ans2,d[n]);

        edg[i][v]/=;

        edg[v][i]/=;

    }

    printf("%d\n",ans2-ans1);

    return ;

}