题目
题解
第一次一遍就AC一道bzoj上的题,虽然是一道水题。。。
我们做一边最短路,求出每个点的dist,然后再做一次类似spfa的操作,求出每个点是否可以用于建图。
在新图上拆点跑一边dinic就好辣。
代码
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
using namespace std;
#define ll long long
const ll inf = 100000000000;
const int maxn = 1000;
struct edge {
int from;
int to;
ll value;
};
vector<edge> edges;
vector<int> G[maxn];
vector<edge> gg[maxn];
void add(int from, int to, ll value) {
gg[from].push_back((edge){from, to, value});
}
void add_edge(int from, int to, ll value) {
edges.push_back((edge){from, to, value});
edges.push_back((edge){to, from, 0});
int m = edges.size();
G[from].push_back(m - 2);
G[to].push_back(m - 1);
}
ll dist[maxn];
int iter[maxn];
int n, m;
ll c[maxn];
int st, t;
void spfa(int s) {
for (int i = 1; i <= n; i++)
dist[i] = inf;
queue<int> q;
bool inq[maxn];
memset(inq, 0, sizeof(inq));
q.push(s);
dist[s] = 0;
inq[s] = 1;
while (!q.empty()) {
int u = q.front();
q.pop();
inq[u] = 0;
for (int i = 0; i < gg[u].size(); i++) {
edge &e = gg[u][i];
if (dist[e.to] > dist[u] + e.value) {
dist[e.to] = dist[u] + e.value;
if (!inq[e.to]) {
q.push(e.to);
inq[e.to] = 1;
}
}
}
}
}
void build_graph(int s, int e) {
queue<int> q;
q.push(s);
bool inq[maxn];
memset(inq, 0, sizeof(inq));
inq[s] = 1;
int vis[maxn];
memset(vis, 0, sizeof(vis));
vis[s] = 1;
add_edge(st, s, inf);
add_edge(e + n, t, inf);
add_edge(s, s + n, inf);
add_edge(e, e + n, inf);
while (!q.empty()) {
int u = q.front();
q.pop();
inq[u] = 0;
for (int i = 0; i < gg[u].size(); i++) {
edge &ee = gg[u][i];
if (dist[ee.to] == dist[u] + ee.value) {
add_edge(u + n, ee.to, inf);
if (!vis[ee.to] && ee.to != e) {
add_edge(ee.to, ee.to + n, c[ee.to]);
vis[ee.to] = 1;
}
if (!inq[ee.to]) {
q.push(ee.to);
inq[ee.to] = 1;
}
}
}
}
}
void bfs(int s) {
memset(dist, -1, sizeof(dist));
dist[s] = 0;
queue<int> q;
q.push(s);
while (!q.empty()) {
int u = q.front();
q.pop();
for (int i = 0; i < G[u].size(); i++) {
edge &e = edges[G[u][i]];
if (e.value > 0 && dist[e.to] == -1) {
dist[e.to] = dist[u] + 1;
q.push(e.to);
}
}
}
}
ll dfs(int s, int t, ll flow) {
if (s == t)
return flow;
for (int &i = iter[s]; i < G[s].size(); i++) {
edge &e = edges[G[s][i]];
if (dist[e.to] > dist[s] && e.value > 0) {
int d = dfs(e.to, t, min(flow, e.value));
if (d > 0) {
e.value -= d;
edges[G[s][i] ^ 1].value += d;
return d;
}
}
}
return 0;
}
ll dinic(int s, int t) {
ll flow = 0;
while (1) {
bfs(s);
if (dist[t] == -1)
return flow;
memset(iter, 0, sizeof(iter));
ll d;
while (d = dfs(s, t, inf))
flow += d;
}
}
int main() {
// freopen("input", "r", stdin);
scanf("%d %d", &n, &m);
for (int i = 1; i <= m; i++) {
int x, y;
ll z;
scanf("%d %d %lld", &x, &y, &z);
add(x, y, z);
add(y, x, z);
}
for (int i = 1; i <= n; i++)
scanf("%lld", &c[i]);
spfa(1);
st = 0, t = n + n + 1;
build_graph(1, n);
ll ans = dinic(st, t);
printf("%lld
", ans);
}
Excited!