Shortest Path(思维，dfs)

Shortest Path

 

 Accepts: 40

 

 Submissions: 610

 Time Limit: 4000/2000 MS (Java/Others)

 

 Memory Limit: 131072/131072 K (Java/Others)

Problem Description

There is a path graph G=(V,E)G=(V,E) with nn vertices. Vertices are numbered from 11 to nn and there is an edge with unit length between iiand i + 1i+1 (1 le i < n)(1≤i<n). To make the graph more interesting, someone adds three more edges to the graph. The length of each new edge is 11.

You are given the graph and several queries about the shortest path between some pairs of vertices.

Input

There are multiple test cases. The first line of input contains an integer TT, indicating the number of test cases. For each test case:

The first line contains two integer nn and mm (1 le n, m le 10^5)(1≤n,m≤10​5​​) -- the number of vertices and the number of queries. The next line contains 6 integers a_1, b_1, a_2, b_2, a_3, b_3a​1​​,b​1​​,a​2​​,b​2​​,a​3​​,b​3​​ (1 le a_1,a_2,a_3,b_1,b_2,b_3 le n)(1≤a​1​​,a​2​​,a​3​​,b​1​​,b​2​​,b​3​​≤n), separated by a space, denoting the new added three edges are (a_1,b_1)(a​1​​,b​1​​), (a_2,b_2)(a​2​​,b​2​​), (a_3,b_3)(a​3​​,b​3​​).

In the next mm lines, each contains two integers s_is​i​​ and t_it​i​​ (1 le s_i, t_i le n)(1≤s​i​​,t​i​​≤n), denoting a query.

The sum of values of mm in all test cases doesn't exceed 10^610​6​​.

Output

For each test cases, output an integer S=(displaystylesum_{i=1}^{m} i cdot z_i) ext{ mod } (10^9 + 7)S=(​i=1​∑​m​​i⋅z​i​​) mod (10​9​​+7), where z_iz​i​​ is the answer for ii-th query.

Sample Input

1
10 2
2 4 5 7 8 10
1 5
3 1

Sample Output

7
题解：最短路，本来完全暴力的最短路，果断超时，最后听了大神的想法，是求加的那三条边的最短路；让每次给的u,v找u到那三个起点的距离与
d[v]的和与v-u找最小值；还是超时，先放着吧，明天再看；
今天看了下发现自己写的很有问题，哪有那么麻烦，直接dfs下就好了，其实就是个思维题，用图论肯定超了。。。
AC代码:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<string>
using namespace std;
const int INF=0x3f3f3f3f;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=1e5+100;
const int MOD=1e9+7;
typedef long long LL;
struct Dot{
    int x,y;
};
Dot dot[3];
int u,v;
LL ans;
int vis[3];
void dfs(int pos,int temp){
    if(temp+abs(v-pos)<ans)ans=temp+abs(v-pos);
    for(int i=0;i<3;i++){
        if(vis[i])continue;
        vis[i]=1;
        dfs(dot[i].y,temp+abs(pos-dot[i].x)+1);
        dfs(dot[i].x,temp+abs(pos-dot[i].y)+1);
        vis[i]=0;
    }
}
int main(){
    int T;
    SI(T);
    int N,M;
    while(T--){
        SI(N);SI(M);
        for(int i=0;i<3;i++)SI(dot[i].x),SI(dot[i].y);
        LL temp=0;
        vis[0]=vis[1]=vis[2]=0;
        for(int i=1;i<=M;i++){
            SI(u);SI(v);
            ans=abs(v-u);
            dfs(u,0);
            temp=(temp+ans*i)%MOD;
        }
        printf("%lld
",temp);
    }
    return 0;
}

我的超时代码：先不说超时了，这样情况都没考虑完全，都没考虑过好几条路的情况；再者这样肯定会超时了。。。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<map>
#include<string>
using namespace std;
const int INF=0x3f3f3f3f;
#define SI(x) scanf("%d",&x)
#define PI(x) printf("%d",x)
#define P_ printf(" ")
#define mem(x,y) memset(x,y,sizeof(x))
const int MAXN=1e5+100;
const int MOD=1e9+7;
int head[MAXN<<1];
int vis[MAXN];
int n,m;
int d1[MAXN],d2[MAXN],d3[MAXN];
int d[MAXN];
struct Node{
    int from,to,next;
};
Node dt[MAXN<<1];
int edgnum;
void add(int u,int v){
    dt[edgnum].from=u;dt[edgnum].to=v;
    dt[edgnum].next=head[u];
    head[u]=edgnum++;
}
void dijkscra(int u){
    mem(vis,0);
    mem(d,INF);
    d[u]=0;
    while(true){
        int k=-1;
        for(int i=1;i<=n;i++){
            if(!vis[i])if(k==-1||d[i]<d[k])k=i;
        }
        if(k==-1)break;
        vis[k]=1;
        for(int i=head[k];i!=-1;i=dt[i].next){
            int u=dt[i].from,v=dt[i].to;
            //printf("%d %d
",u,v);
            d[v]=min(d[v],d[u]+1);
        }
    }
}

/*
int temp;
void dfs(int u,int v,int t){
    if(t>n)return;
    if(u==v){
        temp=min(temp,t);
        return;
    }
    for(int i=head[u];i!=-1;i=dt[i].next){
        dfs(dt[i].to,v,t+1);
    }
}
*/
int main(){
    int T;
    SI(T);
    while(T--){
        SI(n);SI(m);
        int a1,a2,a3,b1,b2,b3;
        edgnum=0;mem(head,-1);
        for(int i=1;i<n;i++)add(i,i+1),add(i+1,i);
        scanf("%d%d%d%d%d%d",&a1,&b1,&a2,&b2,&a3,&b3);
        add(a1,b1);add(b1,a1);add(a2,b2);add(b2,a2);add(a3,b3);add(b3,a3);
        __int64 ans=0;
        dijkscra(a1);
        for(int i=1;i<=n;i++)d1[i]=d[i];
        dijkscra(a2);
        for(int i=1;i<=n;i++)d2[i]=d[i];
        dijkscra(a3);
        for(int i=1;i<=n;i++)d3[i]=d[i];
        
        for(int i=1;i<=m;i++){
            int u,v;
            scanf("%d%d",&u,&v);
            /*dijkscra(u);*/
            int temp=min(min(abs(v-u),abs(u-a1)+d1[v]),min(abs(u-a2)+d2[v],abs(u-a3)+d3[v]));
            //printf("%d
",temp);
            ans=(ans+temp*i)%MOD;
        }
        printf("%I64d
",ans);
    }
    return 0;
}