HDU 5889最小割+最短路

Posted 2020-11-01 tennant

题意：给出一张n个点m条边的无向图，边权均为1，敌人在n点准备走最短路在攻击己方位置1点，现在要在一些边上设置一些路障，给出每条边设置路障的代价，要求用最少的代价设置路障使得敌人必然遇到路障。

这份代码了用到了当前弧优化，尽管我不是很懂。。。
但素不用的话会超时！

#include<stdio.h>
#include<string.h>
#include<queue>
using namespace std;
struct node {
    int from;
    int to;
    int w;
    int next;
}e[160000];
int cur[1500];
int head[1500];
int divv[1500];
int cont, ss, tt;
void add(int from, int to, int w) {
    e[cont].from = from;
    e[cont].w = w;
    e[cont].to = to;
    e[cont].next = head[from];
    head[from] = cont++;
}
int n, m;
int makedivv() {
    memset(divv, 0, sizeof(divv));
    divv[ss] = 1;
    queue<int >s;
    s.push(ss);
    while (!s.empty()) {
        int u = s.front();
        if (u == tt)return 1;
        s.pop();
        for (int i = head[u]; i != -1; i = e[i].next) {
            int w = e[i].w;
            int v = e[i].to;
            if (divv[v] == 0 && w) {
                divv[v] = divv[u] + 1;
                s.push(v);
            }
        }
    }
    return 0;
}
int Dfs(int u, int maxflow, int tt) {
    if (u == tt)return maxflow;
    int ret = 0;
    //当前弧优化
    for (int &i = cur[u]; i != -1; i = e[i].next) {
        int v = e[i].to;
        int w = e[i].w;
        if (divv[v] == divv[u] + 1 && w) {
            int f = Dfs(v, min(maxflow - ret, w), tt);
            e[i].w -= f;
            e[i ^ 1].w += f;
            ret += f;
            if (ret == maxflow)return ret;
        }
    }
    return ret;
}
void Dinic() {
    long long int ans = 0;
    while (makedivv() == 1) {
        memcpy(cur, head, sizeof(head));
        ans += Dfs(ss, 0x3f3f3f3f, tt);
    }
    printf("%lld\n", ans % 100000);
}
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d%d", &n, &m);
        scanf("%d%d", &ss, &tt);
        cont = 0;
        memset(head, -1, sizeof(head));
        for (int i = 0; i < m; i++) {
            int x, y, z;
            scanf("%d%d%d", &x, &y, &z);
            add(x, y, z); add(y, x, 0);
        }
        Dinic();
    }
}