题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5889
解法:求最短路图上的最小割,先在图上源点和终点分别求一遍最短路。再在最短路图上求最小割。
最小割==最大流定理
代码:
#include
#include
#include
#include
#include
using namespace std;
const int MAXN = 10100;//点数的最大值
const int MAXM = 400010;//边数的最大值
const int INF = 0x3f3f3f3f;
struct Edge
{
int to, next, cap, flow;
}edge[MAXM];//注意是MAXM
int tol;
int head[MAXN];
int gap[MAXN], dep[MAXN], pre[MAXN], cur[MAXN];
void init()
{
tol = 0;
memset(head, -1, sizeof(head));
}
//加边,单向图三个参数,双向图四个参数
void addedge(int u, int v, int w, int rw = 0)
{
edge[tol].to = v; edge[tol].cap = w; edge[tol].next = head[u];
edge[tol].flow = 0; head[u] = tol++;
edge[tol].to = u; edge[tol].cap = rw; edge[tol].next = head[v];
edge[tol].flow = 0; head[v] = tol++;
}
//输入参数:起点、终点、点的总数
//点的编号没有影响,只要输入点的总数
int sap(int start, int end, int N)
{
memset(gap, 0, sizeof(gap));
memset(dep, 0, sizeof(dep));
memcpy(cur, head, sizeof(head));
int u = start;
pre[u] = -1;
gap[0] = N;
int ans = 0;
while (dep[start] < N)
{
if (u == end)
{
int Min = INF;
for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])
if (Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
for (int i = pre[u]; i != -1; i = pre[edge[i ^ 1].to])
{
edge[i].flow += Min;
edge[i ^ 1].flow -= Min;
}
u = start;
ans += Min;
continue;
}
bool flag = false;
int v;
for (int i = cur[u]; i != -1; i = edge[i].next)
{
v = edge[i].to;
if (edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u])
{
flag = true;
cur[u] = pre[v] = i;
break;
}
}
if (flag)
{
u = v;
continue;
}
int Min = N;
for (int i = head[u]; i != -1; i = edge[i].next)
if (edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
{
Min = dep[edge[i].to];
cur[u] = i;
}
gap[dep[u]]--;
if (!gap[dep[u]])return ans;
dep[u] = Min + 1;
gap[dep[u]]++;
if (u != start) u = edge[pre[u] ^ 1].to;
}
return ans;
}
int n, m;
int diss[MAXN][MAXN];
bool vis[MAXN];
int d[MAXN];
int d2[MAXN];
void dijkstra(int s)
{
for (int i = 1;i <= n;i++)
{
vis[i] = 0;
if (diss[s][i] == 1)
d[i] = 1;
else
d[i] = INF;
}
d[s] = 0;
while (1)
{
int min = INF, v = -1;
for (int i = 1;i <= n;i++)
if (!vis[i] && d[i]if (v == -1)
break;
vis[v] = 1;
for (int i = 1;i <= n;i++)
if (diss[v][i] == 1)
if (!vis[i] && d[i] > d[v] + 1)
d[i] = 1 + d[v];
}
}
void dijkstra2(int s)
{
for (int i = 1;i <= n;i++)
{
vis[i] = 0;
if (diss[s][i] == 1)
d2[i] = 1;
else
d2[i] = INF;
}
d2[s] = 0;
while (1)
{
int min = INF, v = -1;
for (int i = 1;i <= n;i++)
if (!vis[i] && d2[i]if (v == -1)
break;
vis[v] = 1;
for (int i = 1;i <= n;i++)
if (diss[v][i] == 1)
if (!vis[i] && d2[i] > d2[v] + 1)
d2[i] = 1 + d2[v];
}
}
struct Node
{
int u, v, w;
}node[MAXM];
int main()
{
int T;
scanf("%d", &T);
while (T--)
{
scanf("%d %d", &n, &m);
int u, v, w;
memset(diss, 0, sizeof(diss));
for (int i = 0;i < m;i++)
{
scanf("%d%d%d", &u, &v, &w);
node[i].u = u;
node[i].v = v;
node[i].w = w;
diss[u][v] = diss[v][u] = 1;
}
memset(d, -1, sizeof(d));
memset(d2, -1, sizeof(d2));
dijkstra(1);
dijkstra2(n);
init();
for (int i = 0;i < m;i++)
{
u = node[i].u;
v = node[i].v;
w = node[i].w;
if (d[u] + d2[v] + 1 == d[n] && d[u] !=-1 && d2[v] != -1)
addedge(u, v, w, 0);
if ( d[v] + d2[u] + 1 == d[n] && d[v] != -1 && d2[u] != -1)
addedge(v, u, w, 0);
}
int ans = sap(1, n, n);
printf("%d\n", ans);
}
return 0;
}