Educational Codeforces Round 45 (Rated for Div. 2) F - Flow Control

F - Flow Control

给你一个有向图，要求你给每条边设置流量，使得所有点的流量符合题目给出的要求。

思路：只有在所有点的流量和为0时有解，因为增加一条边的值不会改变所有点的总流量和，

所以我们dfs回溯的时候构造就好了， 其他边设为0。

#include<bits/stdc++.h>
#define LL long long
#define fi first
#define se second
#define mk make_pair
#define pii pair<int, int>

using namespace std;

const int N = 2e5 + ;
const int M = 1e5 + ;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 +;

int head[N], s[N], ans[N], flow[N], tot, n, m, sum;
bool vis[N];
struct Edge {
    int to, id, nx;
} edge[N << ];

void add(int u, int v, int id) {
    edge[tot].to = v;
    edge[tot].id = id;
    edge[tot].nx = head[u];
    head[u] = tot++;
}

void dfs(int u, int p, int id) {
    vis[u] = true;
    for(int i = head[u]; ~i; i = edge[i].nx) {
        int v = edge[i].to;
        if(vis[v]) continue;
        dfs(v, u, i);
    }

    if(flow[u] != s[u]) {
        int ret = abs(s[u] - flow[u]);
        if(flow[u] < s[u]) {
            flow[u] = ;
            flow[p] -= ret;
            if(id & ) {
                ans[edge[id].id] = -ret;
            } else {
                ans[edge[id].id] = ret;
            }
        } else {
            flow[u] = ;
            flow[p] += ret;
            if(id & ) {
                ans[edge[id].id] = ret;
            } else {
                ans[edge[id].id] = -ret;
            }
        }
    }
}

int main() {
    memset(head, -, sizeof(head));
    scanf("%d", &n);
    for(int i = ; i <= n; i++) {
        scanf("%d", &s[i]);
        sum += s[i];
    }

    scanf("%d", &m);
    for(int i = ; i <= m; i++) {
        int u, v; scanf("%d%d", &u, &v);
        add(u, v, i); add(v, u, i);
    }

    if(!sum) {
        dfs(, , -);
        puts("Possible");
        for(int i = ; i <= m; i++) {
            printf("%d\n", ans[i]);
        }
    } else {
        puts("Impossible");
    }
    return ;
}
/*
*/