BZOJ 3931 CQOI 2015 网络吞吐量 最短路+最大流

题目大意

给出一个无向图，求出在这个图上1到n的所有最短路形成的图的最大流。

思路

想让大家叠模板也不带这么懒得吧。。

记得开long long就行了。

CODE

#define _CRT_SECURE_NO_WARNINGS

#include <queue>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define MAXP 10010
#define MAXE 1000010
#define S 0
#define T (MAXP - 1)
#define INF 0x3f3f3f3f3f3f3f3fll
using namespace std;

struct MaxFlow{
    int head[MAXP], total;
    int _next[MAXE], aim[MAXE];
    long long flow[MAXE];

    int deep[MAXP];

    MaxFlow():total(1) {}
    void Add(int x, int y, long long f) {
        _next[++total] = head[x];
        aim[total] = y;
        flow[total] = f;
        head[x] = total;
    }
    void Insert(int x, int y, long long f) {
        Add(x, y, f);
        Add(y, x, 0);
    }
    bool BFS() {
        static queue<int> q;
        while(!q.empty())   q.pop();
        memset(deep, 0, sizeof(deep));
        deep[S] = 1;
        q.push(S);
        while(!q.empty()) {
            int x = q.front(); q.pop();
            for(int i = head[x]; i; i = _next[i])
                if(flow[i] && !deep[aim[i]]) {
                    deep[aim[i]] = deep[x] + 1;
                    q.push(aim[i]);
                    if(aim[i] == T) return true;
                }
        }
        return false;
    }
    long long Dinic(int x, long long f) {
        if(x == T)  return f;
        long long temp = f;
        for(int i = head[x]; i; i = _next[i])
            if(flow[i] && deep[aim[i]] == deep[x] + 1 && temp) {
                long long away = Dinic(aim[i], min(flow[i], temp));
                if(!away)   deep[aim[i]] = 0;
                flow[i] -=away;
                flow[i^1] += away;
                temp -= away;
            }
        return f - temp;
    }
}solver;

struct Edge{
    int x, y, z;

    void Read() {
        scanf("%d%d%d", &x, &y, &z);
    }
}edge[MAXE];

int points, edges;
int head[MAXP], total;
int _next[MAXE], aim[MAXE];
long long length[MAXE];

inline void Add(int x, int y, long long len)
{
    _next[++total] = head[x];
    aim[total] = y;
    length[total] = len;
    head[x] = total;
}

long long l1[MAXP], l2[MAXP];
bool v[MAXP];

void Dijkstra(int start, long long f[])
{
    memset(v, false, sizeof(v));
    f[start] = 0;
    for(int i = 1; i <= points; ++i) {
        long long min_length = INF;
        int p;
        for(int j = 1; j <= points; ++j)
            if(!v[j] && f[j] < min_length)
                min_length = f[j], p = j;
        v[p] = true;
        for(int j = head[p]; j; j = _next[j])
            if(!v[aim[j]])
                f[aim[j]] = min(f[aim[j]], f[p] + length[j]);
    }
}

int main()
{
    cin >> points >> edges;
    for(int i = 1; i <= edges; ++i) {
        edge[i].Read();
        Add(edge[i].x, edge[i].y, edge[i].z);
        Add(edge[i].y, edge[i].x, edge[i].z);
    }
    for(int x, i = 1; i <= points; ++i) {
        scanf("%d", &x);
        solver.Insert(i << 1, i << 1|1, (i == 1 || i == points) ? INF:x);
    }
    memset(l1, 0x3f, sizeof(l1));
    memset(l2, 0x3f, sizeof(l2));
    Dijkstra(1, l1);
    Dijkstra(points, l2);
    long long min_length = l2[1];
    for(int i = 1; i <= edges; ++i) {
        if(l1[edge[i].x] + l2[edge[i].y] + edge[i].z == min_length)
            solver.Insert(edge[i].x << 1|1, edge[i].y << 1, INF);
        if(l2[edge[i].x] + l1[edge[i].x] + edge[i].z == min_length)
            solver.Insert(edge[i].y << 1|1, edge[i].x << 1, INF);
    }
    solver.Insert(S, 2, INF);
    solver.Insert(points << 1|1, T, INF);
    long long max_flow = 0;
    while(solver.BFS())
        max_flow += solver.Dinic(S, INF);
    cout << max_flow << endl;
    return 0;
}