【2019多校第一场补题 / HDU6582】2019多校第一场E题1005Path——最短路径+网络流

HDU6582链接

题意

在一张有向图中，有一个起点和一个终点，你需要删去部分路径，使得起点到终点的最短距离增加（并不要求需要使得距离变成最大值），且删除的路径长度最短。求删去的路径总长为多少

分析

一开始理解错题意了，以为是在保证路径变成最长的路径之后，求删去的路径和最小是多少。然后就自闭了很久，还WA了好几发。后来看到题目中是 longer 而不是 longest 。突然醒悟。直接最短路径 +网络流就行，中间重新建图。
大致的过程是先跑最短路径（我用了SPFA算法，因为当数据量较大时，图为稀疏图，所以用邻接表形式），然后求出起点到每一个点的距离（保存在数组 dist 中）。然后删掉所有的边，对满足下面等式的边进行重建（网络流的边，即同时需要搭建反向的边，只不过流量为0），然后跑网络流（我用了ISAP算法，仍然是邻接表）

$dist[a] - dist[b] = edge[a to b]$
$a to b$ 指代这条边起点为 $a$ 终点为 $b$，且满足 $edge[b to a] = - edge[a to b]$

AC代码

#include <bits/stdc++.h>

using namespace std;

#define MAXN 20100
#define MAXM 20100

bool visited[MAXN];          //标记数组
long long dist[MAXN];        //源点到顶点i的最短距离
long long path[MAXN];        //记录最短路的路径
long long enqueue_num[MAXN]; //记录入队次数
long long vertex_num;        //顶点数
long long edge_num;          //边数
long long source;            //源点

struct Edge
{
    long long to, next, cap, flow;
} edge[MAXM];
long long head[MAXN];
long long tot;
long long gap[MAXN], dep[MAXN], cur[MAXN];
void init()
{
    tot = 0;
    memset(head, -1, sizeof(head));
}

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

bool SPFA()
{
    memset(visited, 0, sizeof(visited));
    memset(enqueue_num, 0, sizeof(enqueue_num));
    for (long long i = 0; i < vertex_num; i++)
    {
        dist[i] = __LONG_LONG_MAX__;
        path[i] = source;
    }

    queue<long long> Q;
    Q.push(source);
    dist[source] = 0;
    visited[source] = true;
    enqueue_num[source]++;
    while (!Q.empty())
    {
        long long u = Q.front();
        Q.pop();
        visited[u] = 0;
        for (long long curnode = head[u]; curnode != -1; curnode = edge[curnode].next)
        {
            if (dist[u] + edge[curnode].cap < dist[edge[curnode].to])
            {
                dist[edge[curnode].to] = dist[u] + edge[curnode].cap;
                path[edge[curnode].to] = u;
                if (!visited[edge[curnode].to])
                {
                    Q.push(edge[curnode].to);
                    enqueue_num[edge[curnode].to]++;
                    if (enqueue_num[edge[curnode].to] >= vertex_num)
                        return false;
                    visited[edge[curnode].to] = 1;
                }
            }
        }
    }
    return true;
}
long long Q[MAXN];
void BFS(long long start, long long end)
{
    memset(dep, -1, sizeof(dep));
    memset(gap, 0, sizeof(gap));
    gap[0] = 1;
    long long front = 0, rear = 0;
    dep[end] = 0;
    Q[rear++] = end;
    while (front != rear)
    {
        long long u = Q[front++];
        for (long long i = head[u]; i != -1; i = edge[i].next)
        {
            long long v = edge[i].to;
            if (dep[v] != -1)
                continue;
            Q[rear++] = v;
            dep[v] = dep[u] + 1;
            gap[dep[v]]++;
        }
    }
}

long long S[MAXN];
long long sap(long long start, long long end, long long N)
{
    BFS(start, end);
    memcpy(cur, head, sizeof(head));
    long long top = 0;
    long long u = start;
    long long ans = 0;
    while (dep[start] < N)
    {
        if (u == end)
        {
            long long Min = __LONG_LONG_MAX__;
            long long inser;
            for (long long i = 0; i < top; i++)
            {
                if (Min > edge[S[i]].cap - edge[S[i]].flow)
                {
                    Min = edge[S[i]].cap - edge[S[i]].flow;
                    inser = i;
                }
            }
            for (long long i = 0; i < top; i++)
            {
                edge[S[i]].flow += Min;
                edge[S[i] ^ 1].flow -= Min;
            }
            ans += Min;
            top = inser;
            u = edge[S[top] ^ 1].to;
            continue;
        }
        bool flag = false;
        long long v;
        for (long long i = cur[u]; i != -1; i = edge[i].next)
        {
            v = edge[i].to;
            if (edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u])
            {
                flag = true;
                cur[u] = i;
                break;
            }
        }
        if (flag)
        {
            S[top++] = cur[u];
            u = v;
            continue;
        }
        long long Min = N;
        for (long long i = head[u]; i != -1; i = edge[i].next)
            if (edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if (!gap[dep[u]])
            return ans;
        dep[u] = Min + 1;
        gap[dep[u]]++;
        if (u != start)
            u = edge[S[--top] ^ 1].to;
    }
    return ans;
}
long long n, m;
int a[MAXN], b[MAXN], c[MAXN];

void reISAP()
{
    init();
    for (int i = 0; i < m; i++)
    {
        if (c[i] == dist[b[i]] - dist[a[i]])
        {
            addedge(a[i], b[i], c[i]);
            addedge(b[i], a[i], 0);
        }
    }
}

int main()
{
#ifdef ACM_LOCAL
    freopen("./in.txt", "r", stdin);
    freopen("./out.txt", "w", stdout);
#endif
    ios::sync_with_stdio(false);
    long long t;
    cin >> t;
    while (t--)
    {
        cin >> n >> m;
        source = 1;
        vertex_num = n + 1;
        init();
        for (long long i = 0; i < m; i++)
        {
            cin >> a[i] >> b[i] >> c[i];
            addedge(a[i], b[i], c[i]);
        }
        if (!SPFA())
        {
            cout << '0' << endl;
            continue;
        }
        reISAP();
        cout << sap(1, n, n) << endl;
    }
    return 0;
}

总结

理解了题意之后感觉就是一道板子题……
人尽皆知**题