题意:有一个恐怖分子要从a点到b点,他可能通过所有线路。在一些点中放置特工,使一定抓到恐怖分子。特工在每个点的价格不一样,求最小总价格。
把每一个点拆点,i->i'流量为权值,连接的点按拆点后连接即可。
gg啊,无向图没看到wa了好几发。。。
链接:hdu- 4289
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <string>
#include <set>
#include <map>
#include <stack>
#include <queue>
#define INF 0x3f3f3f3f
#define ll long long
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 200005;
int n, m;//点数、边数
int sp, tp;//原点、汇点
struct node {
int v, next;
ll cap;
}mp[maxn];
int pre[maxn], cur[maxn];//cur为当前弧优化,dis存储分层图中每个点的层数(即到原点的最短距离),pre建邻接表
ll dis[maxn];
int cnt = 0;
void init() { //不要忘记初始化
cnt = 0;
memset(pre, -1, sizeof(pre));
}
void add(int u, int v, ll w) { //加边
mp[cnt].v = v;
mp[cnt].cap = w;
mp[cnt].next = pre[u];
pre[u] = cnt++;
mp[cnt].v = u;
mp[cnt].cap = 0;
mp[cnt].next = pre[v];
pre[v] = cnt++;
}
bool bfs() { //建分层图
memset(dis, -1, sizeof(dis));
queue<int>q;
while(!q.empty())
q.pop();
q.push(sp);
dis[sp] = 0;
int u, v;
while(!q.empty()) {
u = q.front();
q.pop();
for(int i = pre[u]; i != -1; i = mp[i].next) {
v = mp[i].v;
if(dis[v] == -1 && mp[i].cap > 0) {
dis[v] = dis[u] + 1;
q.push(v);
if(v == tp)
break;
}
}
}
return dis[tp] != -1;
}
ll dfs(int u, ll cap) {//寻找增广路
if(u == tp || cap == 0)
return cap;
ll res = 0, f;
for(int i = cur[u]; i != -1; i = mp[i].next) {//
int v = mp[i].v;
if(dis[v] == dis[u] + 1 && (f = dfs(v, min(cap - res, mp[i].cap))) > 0) {
mp[i].cap -= f;
mp[i ^ 1].cap += f;
res += f;
if(res == cap)
return cap;
}
}
if(!res)
dis[u] = -1;
return res;
}
ll dinic() {
ll ans = 0;
while(bfs()) {
for(int i = 0; i <= 2 * n + 1; i++)
cur[i] = pre[i];
ans += dfs(sp, inf);
}
return ans;
}
int main ()
{
while(~scanf("%d %d", &n, &m)) {
init();
scanf("%d %d", &sp, &tp);
tp += n;
ll p;
for(int i = 1; i <= n; i++) {
scanf("%lld", &p);
add(i, i + n, p);
}
int s, t;
for(int i = 1; i <= m; i++) {
scanf("%d %d", &s, &t);
add(s + n, t, inf);
add(t + n, s, inf);
}
ll res = dinic();
printf("%lld\n", res);
}
return 0;
}