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
題意：由1個總部、若干個城市、中繼站和1些網段構成1個容量網絡。每一個網段是單向的且存在1定帶寬。城市會無窮制的上傳數據，求出該圖中的這樣1些網段：增加這些網段的帶寬可以提高總部收到的總帶寬。
思路：建圖――中繼站和城市都是頂點，網段是邊，總部是匯點。建立超級源點和每一個城市連邊，容量為INF。
要找出這個容量網絡的關鍵邊1，首先要跑1遍最大流，然后在殘留網絡中，分別從源點和匯點進行dfs和染色，如果該邊不是滿流就能夠繼續dfs。最后滿流且兩端點分別具有不同色彩2的邊就是關鍵邊。
另外1定要注意邊的真正條數。應當是3000.
代碼以下：
/*
* 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;//存儲每條網段，省去了編號進程
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;
}
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