专题四1010

题意：

从1走到2有最短路为定值，求最短路有几条

解题思路：先求出最短路，然后搜索求出最短路的条数

感想：明天就考试了。。。

AC源码：

#include<iostream>

#include<queue>

using namespace std;

typedef struct n1

{

int distens,flog;

}node;

node N[1005];

int map[1005][1005],k;

int direct[1005];

void set(int n)

{

int i,j,m,n1,n2,d;

for(i=1;i<=n;i++)

{

for(j=1;j<=n;j++)

{

map[i][j]=-1;

}

N[i].distens=10000000;N[i].flog=0;direct[i]=0;

}

cin>>m;

while(m--)

{

cin>>n1>>n2>>d;

if(map[n1][n2]!=0||map[n1][n2]>d)

map[n1][n2]=map[n2][n1]=d;

}

}

void spfa(int n)

{

queue<int> Q;

int now;

int i;

N[2].distens=0;N[2].flog=1;

Q.push(2);

while(!Q.empty())

{

now=Q.front();

Q.pop();

N[now].flog=0;

//if(q.x==1)break;

for(i=1;i<=n;i++)

if(map[now][i]!=-1)

{

if(N[i].distens>N[now].distens+map[now][i])

{

N[i].distens=N[now].distens+map[now][i];

if(N[i].flog==0)

{

N[i].flog=1;

Q.push(i);

}

}

}

}

}

int DFS(int now,int n)

{

int i;

if(direct[now]>0)

return direct[now];

if(now==2)

{

return 1;

}

for(i=1;i<=n;i++)

if(map[now][i]!=-1&&N[now].distens>N[i].distens)

{

direct[now]+=DFS(i,n);

}

return direct[now];

}

int main()

{

int n;

while(cin>>n&&n)

{

set(n);

spfa(n);

k=DFS(1,n);

cout<<k<<endl;

}

}

Problem J

Time Limit : 2000/1000ms (Java/Other) Memory Limit : 65536/32768K (Java/Other)
Total Submission(s) : 43 Accepted Submission(s) : 12
[align=left]Problem Description[/align]
Jimmy experiences a lot of stress at work these days, especially since his accident made working difficult. To relax after a hard day, he likes to walk home. To make things even nicer, his office is on one side of a forest, and his
house is on the other. A nice walk through the forest, seeing the birds and chipmunks is quite enjoyable. <br>The forest is beautiful, and Jimmy wants to take a different route everyday. He also wants to get home before dark, so he always takes a path to make
progress towards his house. He considers taking a path from A to B to be progress if there exists a route from B to his home that is shorter than any possible route from A. Calculate how many different routes through the forest Jimmy might take. <br>

[align=left]Input[/align]
Input contains several test cases followed by a line containing 0. Jimmy has numbered each intersection or joining of paths starting with 1. His office is numbered 1, and his house is numbered 2. The first line of each test case gives
the number of intersections N, 1 < N ≤ 1000, and the number of paths M. The following M lines each contain a pair of intersections a b and an integer distance 1 ≤ d ≤ 1000000 indicating a path of length d between intersection a and a different intersection
b. Jimmy may walk a path any direction he chooses. There is at most one path between any pair of intersections. <br>

[align=left]Output[/align]
For each test case, output a single integer indicating the number of different routes through the forest. You may assume that this number does not exceed 2147483647<br>

[align=left]Sample Input[/align]

5 6
1 3 2
1 4 2
3 4 3
1 5 12
4 2 34
5 2 24
7 8
1 3 1
1 4 1
3 7 1
7 4 1
7 5 1
6 7 1
5 2 1
6 2 1
0

[align=left]Sample Output[/align]

2
4

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