是I题的数据太水吗？网络流也能过？
对于(u,v),
建立(u,v'),(v', u)容量都是1
(S,u)容量是d[u]
(u',T)容量是d[u]
跑网络流。过了。
#include <bits/stdc++.h>
#define INF 0x7fffffff
using namespace std;
const int maxn = 105;
const int maxm = 205;
struct Edge {
int from, to, cap, flow, next;
Edge(int x=0, int y=0, int c=0, int nex=0)
: from(x), to(y), cap(c), next(nex) {
flow = 0;
}
}edges[100000];
int d[maxn], n, m, sum;
int first[maxn + maxm], cur[maxn + maxm], tots;
int S, T;
void addEdge(int x, int y, int c) {
edges[tots] = Edge(x, y, c, first[x]);
first[x] = tots++;
edges[tots] = Edge(y, x, 0, first[y]);
first[y] = tots++;
}
bool vis[maxn+maxm];
int dep[maxn+maxm];
bool BFS() {
memset(vis, 0, sizeof(vis));
queue<int> Q;
Q.push(0);
dep[0] = 0;
vis[0] = 1;
while (!Q.empty()) {
int x = Q.front(); Q.pop();
for (int i = first[x]; i != -1; i = edges[i].next) {
Edge &e = edges[i];
if (!vis[e.to] && e.cap > e.flow) {
vis[e.to] = 1;
dep[e.to] = dep[x] + 1;
Q.push(e.to);
}
}
}
return vis[T];
}
int DFS(int x, int a) {
if (x == T || a == 0) return a;
int flow = 0, f;
for (int &i = cur[x]; i != -1; i = edges[i].next) {
Edge &e = edges[i];
if (dep[x]+1 == dep[e.to] && (f = DFS(e.to, min(a, e.cap-e.flow))) > 0) {
e.flow += f;
edges[i^1].flow -= f;
flow += f;
a -= f;
if (a == 0) break;
}
}
return flow;
}
int main() {
while (cin >> n >> m) {
sum = 0;
tots = 0;
memset(first, -1, sizeof(first));
for (int i = 1; i <= n; ++i) {
cin >> d[i];
sum += d[i];
}
S = 0;
T = 2*n + 1;
for (int i = 1; i <= n; ++i)
addEdge(S, i, d[i]);
for (int i = 0; i < m; ++i) {
int x, y;
cin >> x >> y;
addEdge(x, y+n, 1);
addEdge(y, x+n, 1);
}
for (int i = n+1; i <= 2*n; ++i)
addEdge(i, T, d[i-n]);
int flow = 0;
while (BFS()) {
for (int i = S; i <= T; ++i)
cur[i] = first[i];
flow += DFS(S, INF);
}
if (flow == sum) printf("Yes\n");
else printf("No\n");
}
return 0;
} 