题目：http://poj.org/problem?id=3469
题意：有双核处理器，有n个任务，给出每个任务在分别在两个处理核心上工作的花费，然后有m行，每行给出两个任务，如果两个任务不在同一个处理核心上工作，那么将有额外的花费。求最小花费
思路：最小割。之前用dinic算法做的，加上当前弧优化6000ms。省赛的时候yjj看到我的最大流板子是 dinic，说到你怎么用这个板子，很容易被卡，学点更快的算法吧。于是回来就学了SAP算法，加当前弧和gap优化，果然更快，2800ms。贴个板子，以备后用
模板一：
这个板子稍微快了一点点。。。但是代码长了很多，因为多了一个bfs函数
#include <iostream>
#include <string>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 20100;
const int INF = 0x3f3f3f3f;
struct edge
{
int to, cap, next;
}g[N*100];
int level[N], cur[N], head[N], que[N], pre[N], gap[N];
int n, m, cnt, nv;
void add_edge(int v, int u, int cap)
{
g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++;
g[cnt].to = v, g[cnt].cap = 0, g[cnt].next = head[u], head[u] = cnt++;
}
void bfs(int t) //从汇点出发反向搜索给为每个点确定距离标号，即每个点到汇点需要经过的最少边数，gap数组储存相同标号的点数
{
memset(level, -1, sizeof level);
memset(gap, 0, sizeof gap);
int st = 0, en = 0;
level[t] = 0;
que[en++] = t;
gap[level[t]]++;
while(st < en)
{
int v = que[st++];
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(level[u] < 0)
{
level[u] = level[v] + 1;
gap[level[u]]++;
que[en++] = u;
}
}
}
}
int sap(int s, int t)
{
bfs(t);
memcpy(cur, head, sizeof head);
//gap[0] = nv;
int v = pre[s] = s, flow = 0, aug = INF;
while(level[s] < nv)
{
bool flag = false;
for(int &i = cur[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(g[i].cap > 0 && level[v] == level[u] + 1)
{
flag = true;
pre[u] = v;
v = u;
aug = min(aug, g[i].cap);
if(v == t) //找到一条增广路
{
flow += aug;
while(v != s) //路径回溯更新残留网络
{
v = pre[v];
g[cur[v]].cap -= aug;
g[cur[v]^1].cap += aug;
}
aug = INF;
}
break;
}
}
if(flag) continue;
int minlevel = nv;
for(int i = head[v]; i != -1; i = g[i].next) //寻找与当前点相连接的点中最小的距离标号
{
int u = g[i].to;
if(g[i].cap > 0 && level[u] < minlevel)
minlevel = level[u], cur[v] = i; //保存弧
}
if(--gap[level[v]] == 0) break; //更新gap数组后如果出现断层，则源点汇点之间必定无流，直接退出
level[v] = minlevel + 1; //更新距离标号
gap[level[v]]++;
v = pre[v]; //转当前点的前驱节点继续寻找允许弧
}
return flow;
}
int main()
{
int a, b, c;
while(~ scanf("%d%d", &n, &m))
{
cnt = 0;
memset(head, -1, sizeof head);
for(int i = 1; i <= n; i++)
{
scanf("%d%d", &a, &b);
add_edge(0, i, a);
add_edge(i, n + 1, b);
}
for(int i = 0; i < m; i++)
{
scanf("%d%d%d", &a, &b, &c);
add_edge(a, b, c);
add_edge(b, a, c);
}
nv = n + 2;
printf("%d\n", sap(0, n + 1));
}
return 0;
} 模板二：
没有bfs函数，初始所有点的距离标号为0，在增广过程中维护距离标号
#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
using namespace std;
const int N = 20100;
const int INF = 0x3f3f3f3f;
struct edge
{
int to, cap, next;
}g[N*100];
int level[N], cur[N], head[N], que[N], pre[N], gap[N];
int n, m, cnt, nv;
void add_edge(int v, int u, int cap)
{
g[cnt].to = u, g[cnt].cap = cap, g[cnt].next = head[v], head[v] = cnt++;
g[cnt].to = v, g[cnt].cap = 0, g[cnt].next = head[u], head[u] = cnt++;
}
int sap(int s, int t)
{
memset(level, 0, sizeof level);
memset(gap, 0, sizeof gap);
memcpy(cur, head, sizeof head);
gap[0] = nv; //初始距离标号均为0，所以gap[0] = nv，即所有点
int v = pre[s] = s, flow = 0, aug = INF;
while(level[s] < nv)
{
bool flag = false;
for(int &i = cur[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(g[i].cap > 0 && level[v] == level[u] + 1)
{
flag = true;
pre[u] = v;
v = u;
aug = min(aug, g[i].cap);
if(v == t)
{
flow += aug;
while(v != s)
{
v = pre[v];
g[cur[v]].cap -= aug;
g[cur[v]^1].cap += aug;
}
aug = INF;
}
break;
}
}
if(flag) continue;
int minlevel = nv;
for(int i = head[v]; i != -1; i = g[i].next)
{
int u = g[i].to;
if(g[i].cap > 0 && level[u] < minlevel)
minlevel = level[u], cur[v] = i;
}
if(--gap[level[v]] == 0) break;
level[v] = minlevel + 1;
gap[level[v]]++;
v = pre[v];
}
return flow;
}
int main()
{
int a, b, c;
while(~ scanf("%d%d", &n, &m))
{
cnt = 0;
memset(head, -1, sizeof head);
for(int i = 1; i <= n; i++)
{
scanf("%d%d", &a, &b);
add_edge(0, i, a);
add_edge(i, n + 1, b);
}
for(int i = 0; i < m; i++)
{
scanf("%d%d%d", &a, &b, &c);
add_edge(a, b, c);
add_edge(b, a, c);
}
nv = n + 2;
printf("%d\n", sap(0, n + 1));
}
return 0;
}
