HDU3416 Marriage Match IV —— 最短路径 + 最大流

题目链接：https://vjudge.net/problem/HDU-3416

 

Marriage Match IV

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4710    Accepted Submission(s): 1412

Problem Description

Do not sincere non-interference。
Like that show, now starvae also take part in a show, but it take place between city A and B. Starvae is in city A and girls are in city B. Every time starvae can get to city B and make a data with a girl he likes. But there are two problems with it, one is starvae must get to B within least time, it's said that he must take a shortest path. Other is no road can be taken more than once. While the city starvae passed away can been taken more than once. 


So, under a good RP, starvae may have many chances to get to city B. But he don't know how many chances at most he can make a data with the girl he likes . Could you help starvae?

 

 

Input

The first line is an integer T indicating the case number.(1<=T<=65)
For each case,there are two integer n and m in the first line ( 2<=n<=1000, 0<=m<=100000 ) ,n is the number of the city and m is the number of the roads.

Then follows m line ,each line have three integers a,b,c,(1<=a,b<=n,0<c<=1000)it means there is a road from a to b and it's distance is c, while there may have no road from b to a. There may have a road from a to a,but you can ignore it. If there are two roads from a to b, they are different.

At last is a line with two integer A and B(1<=A,B<=N,A!=B), means the number of city A and city B.
There may be some blank line between each case.

 

 

Output

Output a line with a integer, means the chances starvae can get at most.

 

 

Sample Input

3 7 8 1 2 1 1 3 1 2 4 1 3 4 1 4 5 1 4 6 1 5 7 1 6 7 1 1 7 6 7 1 2 1 2 3 1 1 3 3 3 4 1 3 5 1 4 6 1 5 6 1 1 6 2 2 1 2 1 1 2 2 1 2

 

 

Sample Output

2 1 1

 

 

Author

starvae@HDU

 

 

Source

HDOJ Monthly Contest – 2010.06.05

 

 

题意：

求有多少条最短路径，要求边不能够重叠（但题目可以有重边，即两个点之间可以有几条边）。

 

题解：

1.先用spfa算法求出最短路径，并且筛掉那些不在最短路径上的边。

2.构建网络流图：对于在最短路径上的边u-->v，在网络流图中构建一条边 u-->v，且边权为1，表明这条边只能被走一次。

3.以起点为源点，以终点为汇点，求出最大流，即为答案。

 

 

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const int INF = 2e9;
const LL LNF = 9e18;
const int mod = 1e9+7;
const int MAXM = 1e5+10;
const int MAXN = 1e3+10;

vector<pair<int,int> >son[MAXN];
int dis1[MAXN], dis2[MAXN], dis[MAXN], inq[MAXN];
void spfa(int start, int N)
{
    memset(inq, 0, sizeof(inq));
    for(int i = 1; i<=N; i++)
        dis[i] = INF;

    queue<int>Q;
    Q.push(start);
    inq[start] = 1;
    dis[start] = 0;
    while(!Q.empty())
    {
        int u = Q.front();
        Q.pop(); inq[u] = 0;
        for(int i = 0; i<son[u].size(); i++)
        {
            int v = son[u][i].first;
            int w = son[u][i].second;
            if(dis[v]>dis[u]+w)
            {
                dis[v] = dis[u]+w;
                if(!inq[v])
                {
                    Q.push(v);
                    inq[v] = 1;
                }
            }
        }
    }
}

struct Edge
{
    int to, next, cap, flow;
}edge[MAXM<<2];
int tot, head[MAXN];
int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];

void init()
{
    tot = 0;
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int w)
{
    edge[tot].to = v; edge[tot].cap = w; edge[tot].flow = 0;
    edge[tot].next = head[u]; head[u] = tot++;
    edge[tot].to = u; edge[tot].cap = 0; edge[tot].flow = 0;
    edge[tot].next = head[v]; head[v] = tot++;
}

int sap(int start, int end, int nodenum)
{
    memset(dep, 0, sizeof(dep));
    memset(gap, 0, sizeof(gap));
    memcpy(cur, head, sizeof(head));
    int u = pre[start] = start, maxflow = 0,aug = INF;
    gap[0] = nodenum;
    while(dep[start]<nodenum)
    {
        loop:
        for(int i = cur[u]; i!=-1; i = edge[i].next)
        {
            int v = edge[i].to;
            if(edge[i].cap-edge[i].flow && dep[u]==dep[v]+1)
            {
                aug = min(aug, edge[i].cap-edge[i].flow);
                pre[v] = u;
                cur[u] = i;
                u = v;
                if(v==end)
                {
                    maxflow += aug;
                    for(u = pre[u]; v!=start; v = u,u = pre[u])
                    {
                        edge[cur[u]].flow += aug;
                        edge[cur[u]^1].flow -= aug;
                    }
                    aug = INF;
                }
                goto loop;
            }
        }
        int mindis = nodenum;
        for(int i = head[u]; i!=-1; i = edge[i].next)
        {
            int v=edge[i].to;
            if(edge[i].cap-edge[i].flow && mindis>dep[v])
            {
                cur[u] = i;
                mindis = dep[v];
            }
        }
        if((--gap[dep[u]])==0)break;
        gap[dep[u]=mindis+1]++;
        u = pre[u];
    }
    return maxflow;
}

int u[MAXM], v[MAXM], w[MAXM];
int main()
{
    int T, n, m;
    scanf("%d", &T);
    while(T--)
    {
        int start, end;
        scanf("%d%d", &n,&m);
        for(int i = 1; i<=m; i++)
            scanf("%d%d%d", &u[i], &v[i], &w[i]);
        scanf("%d%d", &start, &end);

        for(int i = 1; i<=n; i++) son[i].clear();
        for(int i = 1; i<=m; i++) son[u[i]].push_back(make_pair(v[i], w[i]));
        spfa(start, n);
        memcpy(dis1, dis, sizeof(dis1));
        int distance = dis1[end];

        for(int i = 1; i<=n; i++) son[i].clear();
        for(int i = 1; i<=m; i++) son[v[i]].push_back(make_pair(u[i], w[i]));
        spfa(end, n);
        memcpy(dis2, dis, sizeof(dis2));

        init();
        for(int i = 1; i<=m; i++)
            if(dis1[u[i]]+w[i]+dis2[v[i]]==distance)
                add(u[i], v[i], 1);

        int ans = sap(start, end, n);
        printf("%d\n", ans);
    }
}