HDU 4280 Island Transport(网络流模板)

标签：杭电   网络   流量   网络流   

转载请注明出处：http://blog.csdn.net/u012860063

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=4280

2
5 7
3 3
3 0
3 1
0 0
4 5
1 3 3
2 3 4
2 4 3
1 5 6
4 5 3
1 4 4
3 4 2
6 7
-1 -1
0 1
0 2
1 0
1 1
2 3
1 2 1
2 3 6
4 5 5
5 6 3
1 4 6
2 5 5
3 6 4

9
6

题意：有N个岛，M条无向路 每个路有一最大允许的客流量，求从最西的那个岛最多能运用多少乘客到最东的那个岛。

直接上模板：

#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <cstdlib>
#include <climits>
#include <ctype.h>
#include <vector>
#include <queue>
#include <deque>
#include <set>
#include <map>
#include <iostream>
#include <algorithm>
using namespace std;
#define PI acos(-1.0)
#define VM 100047
#define EM 400047
int inf = 0x3f3f3f3f;
struct E
{
    int to, frm, nxt, cap;
}edge[EM];

int head[VM],e,n,m,src,des;
int dep[VM], gap[VM];

void addedge(int cu, int cv, int cw)
{
    edge[e].frm = cu;
    edge[e].to = cv;
    edge[e].cap = cw;
    edge[e].nxt = head[cu];
    head[cu] = e++;
    edge[e].frm = cv;
    edge[e].to = cu;
    edge[e].cap = 0;
    edge[e].nxt = head[cv];
    head[cv] = e++;
}

int que[VM];

void BFS()
{
    memset(dep, -1, sizeof(dep));
    memset(gap, 0, sizeof(gap));
    gap[0] = 1;
    int front = 0, rear = 0;
    dep[des] = 0;
    que[rear++] = des;
    int u, v;
    while (front != rear)
    {
        u = que[front++];
        front = front%VM;
        for (int i=head[u]; i!=-1; i=edge[i].nxt)
        {
            v = edge[i].to;
            if (edge[i].cap != 0 || dep[v] != -1)
                continue;
            que[rear++] = v;
            rear = rear % VM;
            ++gap[dep[v] = dep[u] + 1];
        }
    }
}
int cur[VM],stack[VM];
int Sap()       //sap模板
{
    int res = 0;
    BFS();
    int top = 0;
    memcpy(cur, head, sizeof(head));
    int u = src, i;
    while (dep[src] < n)
    {
        if (u == des)
        {
            int temp = inf, inser = n;
            for (i=0; i!=top; ++i)
                if (temp > edge[stack[i]].cap)
                {
                    temp = edge[stack[i]].cap;
                    inser = i;
                }
            for (i=0; i!=top; ++i)
            {
                edge[stack[i]].cap -= temp;
                edge[stack[i]^1].cap += temp;
            }
            res += temp;
            top = inser;
            u = edge[stack[top]].frm;
        }

        if (u != des && gap[dep[u] -1] == 0)
            break;
        for (i = cur[u]; i != -1; i = edge[i].nxt)
            if (edge[i].cap != 0 && dep[u] == dep[edge[i].to] + 1)
                break;

        if (i != -1)
        {
            cur[u] = i;
            stack[top++] = i;
            u = edge[i].to;
        }
        else
        {
            int min = n;
            for (i = head[u]; i != -1; i = edge[i].nxt)
            {
                if (edge[i].cap == 0)
                    continue;
                if (min > dep[edge[i].to])
                {
                    min = dep[edge[i].to];
                    cur[u] = i;
                }
            }
            --gap[dep[u]];
            ++gap[dep[u] = min + 1];
            if (u != src)
                u = edge[stack[--top]].frm;
        }
    }
    return res;
}

int main()
{
    int T, i;
    scanf("%d", &T);
    while (T--)
    {
        scanf("%d%d", &n, &m);
        int x, y;
        int Min = inf, Max = -inf;
        for (i=1; i<=n; ++i)       //找出起点src 终点des
        {
            scanf("%d%d", &x, &y);
            if (x <= Min)
            {
                src = i;
                Min = x;
            }
            if (x >= Max)
            {
                des = i;
                Max = x;
            }
        }
        e = 0;
        memset(head, -1, sizeof(head));
        int u, v, c;
        for (i=0; i!=m; ++i)
        {
            scanf("%d%d%d", &u, &v, &c);
            addedge(u,v,c);
            addedge(v,u,c);
        }
        int ans = Sap();
        printf("%d\n", ans);
    }
    return 0;
}

HDU 4280 Island Transport(网络流模板),布布扣,bubuko.com

HDU 4280 Island Transport(网络流模板)

标签：杭电   网络   流量   网络流   

原文地址：http://blog.csdn.net/u012860063/article/details/25338163