Internship
Time Limit: 5 Seconds Memory Limit: 32768 KB
CIA headquarter collects data from across the country through its classified network. They have been using optical fibres long before it’s been deployed on any civilian projects. However they are still under a lot pressure recently because the data are growing rapidly. As a result they are considering upgrading the network with new technologies that provide a few times wider bandwidth. In the experiemental stage, they would like to upgrade one segment of their original network in order to see how it performs. And as a CIA intern it’s your responsibility to investigate which segment could actually help increase the total bandwidth the headquarter receives, suppose that all the cities have infinite data to send and the routing algorithm is optimized. As they have prepared the data for you in a few minutes, you are told that they need the result immediately. Well, practically immediately.
Input
Input contains multiple test cases. First line of each test case contains three integers n, m and l, they represent the number of cities, the number of relay stations and the number of segments. Cities will be referred to as integers from 1 to n, while relay stations use integers from n+1 to n+m. You can saves assume that n + m <= 100, l <= 1000 (all of them are positive). The headquarter is identified by the integer 0.
The next l lines hold a segment on each line in the form of a b c, where a is the source node and b is the target node, while c is its bandwidth. They are all integers where a and b are valid identifiers (from 0 to n+m). c is positive. For some reason the data links are all directional.
The input is terminated by a test case with n = 0. You can safely assume that your calculation can be housed within 32-bit integers.
Output
For each test print the segment id’s that meets the criteria. The result is printed in a single line and sorted in ascending order, with a single space as the separator. If none of the segment meets the criteria, just print an empty line. The segment id is 1 based not 0 based.
Sample Input
2 1 3
1 3 2
3 0 1
2 0 1
2 1 3
1 3 1
2 3 1
3 0 2
0 0 0
Sample Output
2 3
题意：由1个总部、若干个城市、中继站和1些网段构成1个容量网络。每一个网段是单向的且存在1定带宽。城市会无穷制的上传数据，求出该图中的这样1些网段：增加这些网段的带宽可以提高总部收到的总带宽。
思路：建图――中继站和城市都是顶点，网段是边，总部是汇点。建立超级源点和每一个城市连边，容量为INF。
要找出这个容量网络的关键边1，首先要跑1遍最大流，然后在残留网络中，分别从源点和汇点进行dfs和染色，如果该边不是满流就能够继续dfs。最后满流且两端点分别具有不同色彩2的边就是关键边。
另外1定要注意边的真正条数。应当是3000.
代码以下：
/*
* ID: j.sure.1
* PROG:
* LANG: C++
*/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <set>
#include <string>
#include <climits>
#include <iostream>
#define PB push_back
#define LL long long
using namespace std;
const int INF = 0x3f3f3f3f;
const double eps = 1e⑻;
/****************************************/
const int N = 100 + 5, M = 3e3 + 5;
int n, m, l, tot, src, sink;
struct Edge {
int u, v, w, next;
Edge(){}
Edge(int _u, int _v, int _w, int _next):
u(_u), v(_v), w(_w), next(_next){}
}e[M];
int head[N], cur[N], s[N], lev[N];
int line[M];
bool vis1[N], vis2[N];
void init()
{
memset(head, -1, sizeof(head));
memset(vis1, 0, sizeof(vis1));
memset(vis2, 0, sizeof(vis2));
tot = 0;
}
void add(int u, int v, int w)
{
e[tot] = Edge(u, v, w, head[u]);
head[u] = tot++;
e[tot] = Edge(v, u, 0, head[v]);
head[v] = tot++;
}
bool bfs()
{
queue <int> q;
memset(lev, -1, sizeof(lev));
lev[src] = 0;
q.push(src);
while(!q.empty()) {
int u = q.front(); q.pop();
for(int i = head[u]; ~i; i = e[i].next) {
int v = e[i].v;
if(e[i].w && lev[v] == -1) {
lev[v] = lev[u] + 1;
q.push(v);
if(v == sink) return true;
}
}
}
return false;
}
int Dinic()
{
int ret = 0;
while(bfs()) {
memcpy(cur, head, sizeof(cur));
int u = src, top = 0;
while(1) {
if(u == sink) {
int mini = INF, loc;
for(int i = 0; i < top; i++) {
if(mini > e[s[i]].w) {
mini = e[s[i]].w;
loc = i;
}
}
for(int i = 0; i < top; i++) {
e[s[i]].w -= mini;
e[s[i]^1].w += mini;
}
ret += mini;
top = loc;
u = e[s[top]].u;
}
int &i = cur[u];
for(; ~i; i = e[i].next) {
int v = e[i].v;
if(e[i].w && lev[v] == lev[u] + 1) break;
}
if(~i) {
s[top++] = i;
u = e[i].v;
}
else {
if(!top) break;
lev[u] = -1;
u = e[s[--top]].u;
}
}
}
return ret;
}
void dfs1(int u)
{
vis1[u] = 1;
for(int i = head[u]; ~i; i = e[i].next) {
int v = e[i].v;
if(!vis1[v] && e[i].w) dfs1(v);
}
}
void dfs2(int u)
{
vis2[u] = 1;
for(int i = head[u]; ~i; i = e[i].next) {
int v = e[i].v;
if(!vis2[v] && e[i^1].w) dfs2(v);
}
}
int main()
{
#ifdef J_Sure
freopen("000.in", "r", stdin);
//freopen("999.out", "w", stdout);
#endif
while(scanf("%d%d%d", &n, &m, &l), n) {
int u, v, w;
init();
src = n+m+1; sink = 0;
for(int i = 1; i <= n; i++) {
add(src, i, INF);
}
for(int i = 0; i < l; i++) {
scanf("%d%d%d", &u, &v, &w);
line[i] = tot;//存储每条网段，省去了编号进程
add(u, v, w);
}
Dinic();
dfs1(src); dfs2(sink);
bool fir = false;
for(int i = 0; i < l; i++) {
if(!e[line[i]].w && vis1[e[line[i]].u] && vis2[e[line[i]].v]) {
if(fir) printf(" "); fir = true;
printf("%d", i+1);
}
}
puts("");
}
return 0;
}