题目链接：http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2205
题意：
给定一个有向网络，每条边均有一个容量。问是否存在一个从点1->n，流量为C的流。如果不存在，是否可以恰好修改一条弧的容量，使得这样的流存在？
第一行输入 N ,E, C (n个点, e条边）
下面e行u v cap 表示边
若最大流>=c 直接输出 possible
若要修改 则输出possible option: 和弧的列表（左端点小到大排序，再右端点小到大排序）
若不存在 则输出 not possible
思路：
先求一次最大流，>=c直接输出答案
否则，修改边（修改有意义的边一定是满流的（即最小割中的边））
一个优化：求完最大流后把流量留着，以后每次在它的基础上增广
#include<stdio.h>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<string.h>
#include<iostream>
#include<math.h>
#include<queue>
#include<map>
using namespace std;
#define ll long long
#define N 1005
#define M 20000
#define inf 536870912
//N为点数 M为边数
struct Edge{
int from, to, cap, nex;
}edge[M*10];//注意这个一定要够大 不然会re 还有反向弧
int head[N], edgenum;
void addedge(int u, int v, int cap){
Edge E = { u, v, cap, head[u]};
edge[ edgenum ] = E;
head[u] = edgenum ++;
Edge E2= { v, u, 0, head[v]};
edge[ edgenum ] = E2;
head[v] = edgenum ++;
}
int sign[N], s, t;
bool BFS(int from, int to){
memset(sign, -1, sizeof(sign));
sign[from] = 0;
queue<int>q;
q.push(from);
while( !q.empty() ){
int u = q.front(); q.pop();
for(int i = head[u]; i!=-1; i = edge[i].nex)
{
int v = edge[i].to;
if(sign[v]==-1 && edge[i].cap)
{
sign[v] = sign[u] + 1, q.push(v);
if(sign[to] != -1)return true;
}
}
}
return false;
}
int Stack[M*4], top, cur[M*4];
struct Change{
int sign, val;
Change(int a=0,int b=0):sign(a),val(b){}
}haha[M*10];
int hahatop;
void release(){
for(int i = 0; i < hahatop; i++)
edge[haha[i].sign].cap += haha[i].val;
}
int dinic(){
hahatop = 0;
int ans = 0;
while( BFS(s, t) )
{
memcpy(cur, head, sizeof(head));
int u = s; top = 0;
while(1)
{
if(u == t)
{
int flow = inf, loc;//loc 表示 Stack 中 cap 最小的边
for(int i = 0; i < top; i++)
if(flow > edge[ Stack[i] ].cap)
{
flow = edge[Stack[i]].cap;
loc = i;
}
for(int i = 0; i < top; i++)
{
edge[ Stack[i] ].cap -= flow;
edge[Stack[i]^1].cap += flow;
haha[hahatop++] = Change(Stack[i], flow);
haha[hahatop++] = Change(Stack[i]^1, -flow);
}
ans += flow;
top = loc;
u = edge[Stack[top]].from;
}
for(int i = cur[u]; i!=-1; cur[u] = i = edge[i].nex)//cur[u] 表示u所在能增广的边的下标
if(edge[i].cap && (sign[u] + 1 == sign[ edge[i].to ]))break;
if(cur[u] != -1)
{
Stack[top++] = cur[u];
u = edge[ cur[u] ].to;
}
else
{
if( top == 0 )break;
sign[u] = -1;
u = edge[ Stack[--top] ].from;
}
}
}
return ans;
}
int n, e, c;
struct Ans{
int f, t;
bool operator<(const Ans a)const{
if(a.f == f)return a.t>t;
return a.f>f;
}
Ans(int a=0,int b=0):f(a),t(b){}
}ans[M*2];
int main(){
int i, j, u, v, cap, Cas = 1;
while(scanf("%d %d %d",&n,&e,&c), n+e+c){
memset(head, -1, sizeof(head)); edgenum = 0;
while(e--){
scanf("%d %d %d",&u,&v,&cap);
addedge(u,v,cap);
}
printf("Case %d: ", Cas++);
s = 1; t = n;
int maxflow = dinic();
if(maxflow >= c){puts("possible");continue;}
int anstop = 0;
for(i = 0; i < edgenum; i+=2)if(edge[i].cap == 0){
edge[i].cap += c;
if(dinic()+maxflow>=c)ans[anstop++] = Ans(edge[i].from,edge[i].to);
release();
edge[i].cap = 0;
}
if(anstop==0)puts("not possible");
else
{
sort(ans, ans+anstop);
printf("possible option");
for(i=0;i<anstop;i++)printf("%c(%d,%d)",i?‘,‘:‘:‘,ans[i].f,ans[i].t);
puts("");
}
}
return 0;
}
/*
4 4 5
1 2 5
1 3 5
2 4 5
3 4 5
4 4 5
1 2 1
1 3 5
2 4 5
3 4 1
4 4 5
1 2 1
1 3 1
2 4 1
3 4 1
0 0 0
*/
Uva 11248 求只修改一条边的流量使得最大流>c 的所有可行边
原文：http://blog.csdn.net/acmmmm/article/details/19765117