HDU4309-Seikimatsu Occult Tonneru(最大流)
Seikimatsu Occult Tonneru
Time Limit: 12000/6000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1741 Accepted Submission(s): 438
There are N cities in Heaven Empire, where people live, with 3 kinds of directed edges connected with each other. The 1st kind of edges is one of Great Tunnels( no more than 20 tunnels) where a certain number of people can hide here; people can also go through
one tunnel from one city to another. The 2nd kind of edges is the so-called Modern Road, which can only let people go through. The 3rd kind of edges is called Ancient Bridge and all the edges of this kind have different names from others, each of which is
named with one of the twelve constellations( such as Libra, Leo and so on); as they were build so long time ago, they can be easily damaged by one person's pass. Well, for each bridge, you can spend a certain deal of money to fix it. Once repaired, the 3rd
kind of edges can let people pass without any limitation, namely, you can use one bridge to transport countless people. As for the former two kinds of edges, people can initially go through them without any limitation.
We want to shelter the most people with the least money.
Now please tell me the largest number of people who can hide in the Tunnels and the least money we need to spend to realize our objective.
The first line, two integers: N (N<=100), m (m<=1000). They stands for the number of cities and edges.
The next line, N integers, which represent the number of people in the N cities.
Then m lines, four intergers each: u, v, w, p (1<=u, v<=N, 0<=w<=50). A directed edge u to v, with p indicating the type of the edge: if it is a Tunnel then p < 0 and w means the maximum number people who can hide in the the tunnel; if p == 0 then it is a Modern
Road with w means nothing; otherwise it is an Ancient Bridge with w representing the cost of fixing the bridge. We promise there are no more than one edge from u to v.
4 4
2 1 1 0
1 2 0 0
1 3 0 0
2 4 1 -1
3 4 3 -1 4 4
2 1 1 0
1 2 0 0
1 3 3 1
2 4 1 -1
3 4 3 -1
4 0
4 3
#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <string>
#include <algorithm>
#include <queue>
using namespace std;
const int maxn = 100+10;
const int maxm = 1000+10;
const int inf = 1<<25;
int n,m,nume;
struct edge{
int v,f,nxt;
};
struct build{
int u,v,w;
build(int u,int v,int w):u(u),v(v),w(w){}
};
vector<build> vb[3];
edge e[maxm];
int head[maxn];
int city[maxn];
void addedge(int u,int v,int c){
e[++nume].nxt = head[u];
e[nume].v = v;
e[nume].f = c;
head[u] = nume;
e[++nume].nxt = head[v];
e[nume].v = u;
e[nume].f = 0;
head[v] = nume;
}
void init(){
memset(head,0,sizeof head);
nume = 1;
} queue<int> que;
bool vis[maxn];
int dist[maxn];
int src,sink; void bfs(){
memset(dist,0,sizeof dist);
while(!que.empty()) que.pop();
vis[src] = true;
que.push(src);
while(!que.empty()){
int u = que.front();
que.pop();
for(int i = head[u]; i ; i = e[i].nxt){
if(e[i].f && !vis[e[i].v]){
que.push(e[i].v);
vis[e[i].v] = 1;
dist[e[i].v] = dist[u]+1;
}
}
}
} int dfs(int u,int delta){
if(u== sink) return delta;
else{
int ret = 0;
for(int i = head[u]; delta&&i; i = e[i].nxt){
if(e[i].f && dist[e[i].v] == dist[u]+1){
int dd = dfs(e[i].v,min(e[i].f,delta));
e[i].f -= dd;
e[i^1].f += dd;
delta -= dd;
ret += dd;
}
}
return ret;
}
} int maxflow(){
int ret = 0;
while(true){
memset(vis,0,sizeof vis);
bfs();
if(!vis[sink]) return ret;
ret += dfs(src,inf);
} }
int main(){ while(~scanf("%d%d",&n,&m)){
src = 0;
sink = n+1;
for(int i = 0; i < 3; i++) vb[i].clear();
for(int i = 1; i <= n; i++) scanf("%d",&city[i]);
bool flag = 0;
while(m--){
int a,b,c,d;
scanf("%d%d%d%d",&a,&b,&c,&d);
if(d==0){
vb[1].push_back(build(a,b,c));//现代桥
}
else if(d<0){
vb[0].push_back(build(a,b,c));//隧道
flag = 1;
}else{
vb[2].push_back(build(a,b,c));//古代桥
}
}
if(!flag){
printf("Poor Heaven Empire\n");
continue;
}
int d = vb[2].size();
int maxf = -1,minc = inf;
for(int i = 0; i < (1<<d); i++){
init();
int tc = 0;
for(int j = 0; j < d; j++){
if(i&(1<<j)){
addedge(vb[2][j].u,vb[2][j].v,inf);
tc += vb[2][j].w;
}else{
addedge(vb[2][j].u,vb[2][j].v,1);
}
}
for(int i = 1; i <= n; i++){
addedge(0,i,city[i]);
}
for(int i = 0; i < vb[1].size(); i++){
addedge(vb[1][i].u,vb[1][i].v,inf);
}
for(int i = 0; i < vb[0].size(); i++){
addedge(vb[0][i].u,vb[0][i].v,inf);
addedge(vb[0][i].u,n+1,vb[0][i].w);
}
int tm = maxflow();
if(tm > maxf){
maxf = tm;
minc = tc;
}
if(tm==maxf && minc > tc){
minc = tc;
} }
if(maxf==-1){
printf("Poor Heaven Empire\n");
}else{
printf("%d %d\n",maxf,minc);
}
}
return 0;
}
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