题目链接：HDU-4280 Island Transport

题意

给出$n$个岛屿$m$条双向航道，岛屿以直角坐标系上的坐标形式给出，每条航道在单位时间有运输量上限，问单位时间内从$x$轴最左边的岛屿到最右边的岛屿最大的运输量。

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思路

最大流问题，最左边的岛屿为源点，最右边的岛屿为汇点，按所给航道连边即可，由于是无向边，所以不需要再建容量为0的反向边。$n,m$很大，要用效率较高的Dinic算法求解最大流。

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代码实现

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#define RG register
#define getc() (S == T && (T = (S = B) + fread(B, 1, 1 << 15, stdin), S == T) ? EOF : *S++)
using std::queue;
const int INF = 0x3f3f3f3f, N = 100010, M = 200020;
int head[N], d[N];
int s, t, tot, maxflow;
struct Edge
{
    int to, cap, nex;
} edge[M];
queue<int> q;
char B[1 << 15], *S = B, *T = B;
inline int read() {
    RG char ch; RG int x = 0; RG bool m = 0;
    while (ch = getc(), (ch < '0' || ch > '9') && ch != '-') ;
    ch == '-' ? m = 1 : x = ch - '0';
    while (ch = getc(), ch >= '0' && ch <= '9') x = x * 10 + ch - '0';
    return m ? -x : x;
}
void add(int x, int y, int z) {
    edge[++tot].to = y, edge[tot].cap = z, edge[tot].nex = head[x], head[x] = tot;
    //edge[++tot].to = x, edge[tot].cap = 0, edge[tot].nex = head[y], head[y] = tot;
}
bool bfs() {
    memset(d, 0, sizeof(d));
    while (q.size()) q.pop();
    q.push(s); d[s] = 1;
    while (q.size()) {
        int x = q.front(); q.pop();
        for (int i = head[x]; i; i = edge[i].nex) {
            int v = edge[i].to;
            if (edge[i].cap && !d[v]) {
                q.push(v);
                d[v] = d[x] + 1;
                if (v == t) return true;
            }
        }
    }
    return false;
}
int dinic(int x, int flow) {
    if (x == t) return flow;
    int rest = flow, k;
    for (int i = head[x]; i && rest; i = edge[i].nex) {
        int v = edge[i].to;
        if (edge[i].cap && d[v] == d[x] + 1) {
            k = dinic(v, std::min(rest, edge[i].cap));
            if (!k) d[v] = 0;
            edge[i].cap -= k;
            edge[i^1].cap += k;
            rest -= k;
        }
    }
    return flow - rest;
}
void init() {
    tot = 1, maxflow = 0;
    memset(head, 0, sizeof(head));
}

int main() {
    int T = read();
    while (T--) {
        init();
        int n = read(), m = read();
        int xx, yy, lx = INF, rx = -INF;
        for (int i = 0; i < n; i++) {
            xx = read(), yy = read();
            if (xx < lx) lx = xx, s = i + 1;
            if (xx > rx) rx = xx, t = i + 1;
        }
        for (int i = 0, u, v, z; i < m; i++) {
            u = read(), v = read(), z = read();
            add(u, v, z);
            add(v, u, z);
        }
        while (bfs()) maxflow += dinic(s, INF);
        printf("%d\n", maxflow);
    }
    return 0;
}

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