P1343 地震逃生(网络流入门)

传送门

题目大意:学生们同时逃生,有多条路径选择,路径容量有限,一批一批逃生,问最少需要多少批才能逃生完成。

思路:最大流入门题,都根本不需要考虑如何建图,提单第一题,顺利过关。

AC代码:

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 100005;
const int inf = 0x3f3f3f3f;
struct edge {
    int f, t, nxt;
    ll flow;
}e[maxn];
int hd[maxn], tot=1;
void add(int f, int t,ll flow) {
    e[++tot] = { f,t,hd[f],flow };
    hd[f] = tot;
}
int n, m, x;
int s, d;
int dep[maxn], cur[maxn];
bool bfs() {//找增广路
    memset(dep, 0, sizeof(dep));
    dep[s] = 1;
    queue<int>Q;
    Q.push(s);
    while (!Q.empty()) {
        int u = Q.front(); Q.pop();
        for (int i = hd[u]; i; i = e[i].nxt) {
            int v = e[i].t, flow = e[i].flow;
            if (flow > 0 && !dep[v]) {
                dep[v] = dep[u] + 1;
                if (v == d)return true;
                Q.push(v);
            }
        }
    }
    return false;
}
ll dfs(int u, ll flow) {
    if (u == d)return flow;
    ll last = flow;
    for (int i = cur[u]; i; i = e[i].nxt) {
        cur[u] = i;//当前弧优化
        int v = e[i].t;ll flow = e[i].flow;
        if (flow > 0 && dep[v] == dep[u] + 1) {
            ll tmp = dfs(v, min(last, flow));
            last -= tmp;
            e[i].flow -= tmp;
            e[i ^ 1].flow += tmp;
            if (last == 0)break;//流量没了直接退出循环,与当前弧优化对应
        }
    }
    if (last == flow)dep[u] = 0;//从当前点一点流量没流到终点(未找到增广路),炸点优化
    return flow - last;//返回剩余流量
}
ll dinic() {
    ll maxflow = 0;
    while (bfs()) {
        memcpy(cur, hd, sizeof(hd));
        maxflow += dfs(s, inf);
    }
    return maxflow;
}
int main() {
    //freopen("test.txt", "r", stdin);
    scanf("%d%d%d", &n, &m, &x);
    for (int i = 1; i <= m; i++) {
        int a, b, c; scanf("%d%d%d", &a, &b, &c);
        add(a, b, c);
        add(b, a, 0);
    }
    s = n + 1, d = n + 2;
    add(s, 1, inf);
    add(n, d, inf);
    ll ans = dinic();
    if (!ans) {
        printf("Orz Ni Jinan Saint Cow!\n");
    }else printf("%lld %lld\n", ans, (x + ans-1) / ans);
    return 0;
}