洛谷P3254 圆桌问题(最大流)

题意

$m$个不同单位代表参加会议，第$i$个单位有$r_i$个人

$n$张餐桌，第$i$张可容纳$c_i$个代表就餐

同一个单位的代表需要在不同的餐桌就餐

问是否可行，要求输出方案

Sol

比较zz的最大流

从$S$向$1-m$连流量为$r_i$的边

从$m + 1$向$m + n$连流量为$c_i$的边

从$1-m$向$m + 1$到$m + n$中的每个点连流量为$1$的边

跑最大流即可

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
const int MAXN = 1e5 + 10, INF = 1e9 + 10;
inline int read() {char c = getchar(); int x = 0, f = 1;while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();return x * f;
}
int M, N, S, T;
int r[MAXN], c[MAXN];
struct Edge {int u, v, f, nxt;
}E[MAXN];
int head[MAXN], cur[MAXN], num;
inline void add_edge(int x, int y, int f) {E[num] = (Edge){x, y, f, head[x]};head[x] = num++;
}
inline void AddEdge(int x, int y, int z) {add_edge(x, y, z);add_edge(y, x, 0);
}
int sum = 0, deep[MAXN];
bool BFS() {queue<int> q; q.push(S);memset(deep, 0, sizeof(deep)); deep[S] = 1;while(!q.empty()) {int p = q.front(); q.pop();for(int i = head[p]; i != -1; i = E[i].nxt) {int to = E[i].v;if(!deep[to] && E[i].f) {deep[to] = deep[p] + 1;q.push(to);}}}return deep[T] > 0;
}
int DFS(int x, int flow) {if(x == T) return flow;int ansflow = 0;for(int &i = cur[x]; i != -1; i = E[i].nxt) {int to = E[i].v;if(deep[to] == deep[x] + 1 && E[i].f) {int nowflow = DFS(to, min(flow, E[i].f));E[i].f -= nowflow; E[i ^ 1].f += nowflow;ansflow += nowflow; flow -= nowflow;if(flow <= 0) break;}}return ansflow;
}
int Dinic() {int ans = 0;while(BFS()) {memcpy(cur, head, sizeof(head));ans += DFS(S, INF);}return ans;
}
int main() {memset(head, -1, sizeof(head));M = read(); N = read(); S = 0; T = M + N + 1;for(int i = 1; i <= M; i++) r[i] = read(), AddEdge(S, i, r[i]), sum += r[i];for(int i = 1; i <= N; i++) c[i] = read(), AddEdge(i + M, T, c[i]);for(int i = 1; i <= M; i++)for(int j = 1; j <= N; j++)AddEdge(i, j + M, 1);if(Dinic() >= sum) printf("1\n");else {printf("0"); return 0;} for(int x = 1; x <= M; x++) {for(int i = head[x]; i != -1; i = E[i].nxt) if(E[i].f == 0)printf("%d ", E[i].v - M);puts("");}return 0;
}