HDU 3416 Marriage IV(the shortest path and maxflow)
Description
求最短路有几种方案
（不同的两种方案所经过的边完全不同！）
Analyze
以 A 为起点跑一遍spfa， 得到 dis_a[]
以 B 为起点跑一遍spfa， 得到 dis_b[]
对于边(u, v, w)，若满足dis_a[u] + dis_b[v] + w == dis_a[B]
则这条边是最短路的可选边(可以作为最短路上的一条边)
找出所有的可选边，设其容量为1，A到B的最大流即完全不同的最短路方案数
结论：图论相关的题目，求有几种方案：将确定边设容量为1，求最大流即可
Code
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define Begin cout << "Check Begin---------------\n";
#define END cout << "Check End-----------------\n";
#define Exit cout << "Exit---------------------\n";
const int INF = 0x3f3f3f3f;
const int N = int(1e3) + 5;
const int M = int(1e5) + 5;
int n, m, start, over;
int cnt_a, cnt_b, cnt_c;
struct edge
{
int to, w, pre;
} a[M], b[M], c[2 * M];
int dis_a[N], dis_b[N];
bool vis_a[N], vis_b[N];
int head_a[N], head_b[N], head_c[N];
void init()
{
cnt_a = 0, cnt_b = 0, cnt_c = 0;
for(int i = 0; i <= n; ++i)
{
dis_a[i] = dis_b[i] = INF;
vis_a[i] = vis_b[i] = 0;
head_a[i] = head_b[i] = head_c[i] = -1;
}
}
void add(int & cnt, edge * e, int * head, int from, int to, int w)
{
e[cnt].to = to;
e[cnt].w = w;
e[cnt].pre = head[from];
head[from] = cnt++;
}
void spfa(int s, int * head, edge * e, int * dis, bool * vis)
{
queue<int> q;
q.push(s);
dis[s] = 0;
vis[s] = 1;
while(!q.empty())
{
int now = q.front();
q.pop();
vis[now] = 0;
for(int i = head[now]; ~i; i = e[i].pre)
{
int to = e[i].to;
if(dis[to] > dis[now] + e[i].w)
{
dis[to] = dis[now] + e[i].w;
if(!vis[to])
q.push(to), vis[to] = 1;
}
}
}
}
int maxflow, deep[N], cur[N];
bool BFS(int s, int t)
{
memset(deep, INF, sizeof(deep));
for(int i = 0; i <= n; ++i)
cur[i] = head_c[i];
deep[s] = 0;
queue<int> q;
q.push(s);
while(!q.empty())
{
int now = q.front();
q.pop();
for(int i = head_c[now]; ~i; i = c[i].pre)
if(deep[ c[i].to ] == INF && c[i].w)
{
deep[ c[i].to ] = deep[now] + 1;
q.push(c[i].to);
}
}
if(deep[t] < INF)
return 1;
return 0;
}
int DFS(int now, int t, int limit)
{
if(!limit || now == t)
return limit;
int flow = 0, f;
for(int i = cur[now]; ~i; i = c[i].pre)
{
cur[now] = i;
if(deep[ c[i].to ] == deep[now] + 1)
if(f = DFS(c[i].to, t, min(limit, c[i].w)))
{
flow += f;
limit -= f;
c[i].w -= f;
c[i ^ 1].w += f;
if(!limit)
break;
}
}
return flow;
}
void Dinic(int s, int t)
{
int tem;
while(BFS(s, t))
while(tem = DFS(s, t, INF))
maxflow += tem;
}
int main()
{
int _;
scanf("%d", &_);
while(_--)
{
scanf("%d %d", &n, &m);
init();
int u[M], v[M], w[M];
for(int i = 1; i <= m ; ++i)
{
scanf("%d %d %d", &u[i], &v[i], &w[i]);
if(u[i] == v[i])
continue;
add(cnt_a, a, head_a, u[i], v[i], w[i]);
add(cnt_b, b, head_b, v[i], u[i], w[i]);
}
scanf("%d %d", &start, &over);
spfa(start, head_a, a, dis_a, vis_a);
spfa(over, head_b, b, dis_b, vis_b);
for(int i = 1; i <= m; ++i)
if(u[i] != v[i] && dis_a[ u[i] ] + dis_b[ v[i] ] + w[i] == dis_a[over])
{
add(cnt_c, c, head_c, u[i], v[i], 1);
add(cnt_c, c, head_c, v[i], u[i], 0);
}
maxflow = 0;
Dinic(start, over);
cout << maxflow << '\n';
}
return 0;
}
- 1
- 2
- 3
- 4
- 5
- 6
- 7
- 8
- 9
- 10
- 11
- 12
- 13
- 14
- 15
- 16
- 17
- 18
- 19
- 20
- 21
- 22
- 23
- 24
- 25
- 26
- 27
- 28
- 29
- 30
- 31
- 32
- 33
- 34
- 35
- 36
- 37
- 38
- 39
- 40
- 41
- 42
- 43
- 44
- 45
- 46
- 47
- 48
- 49
- 50
- 51
- 52
- 53
- 54
- 55
- 56
- 57
- 58
- 59
- 60
- 61
- 62
- 63
- 64
- 65
- 66
- 67
- 68
- 69
- 70
- 71
- 72
- 73
- 74
- 75
- 76
- 77
- 78
- 79
- 80
- 81
- 82
- 83
- 84
- 85
- 86
- 87
- 88
- 89
- 90
- 91
- 92
- 93
- 94
- 95
- 96
- 97
- 98
- 99
- 100
- 101
- 102
- 103
- 104
- 105
- 106
- 107
- 108
- 109
- 110
- 111
- 112
- 113
- 114
- 115
- 116
- 117
- 118
- 119
- 120
- 121
- 122
- 123
- 124
- 125
- 126
- 127
- 128
- 129
- 130
- 131
- 132
- 133
- 134
- 135
- 136
- 137
- 138
- 139
- 140
- 141
- 142
- 143
- 144
- 145
- 146
- 147
- 148
- 149
- 150
- 151
- 152
- 153
- 154
- 155
- 156
- 157
- 158
- 159
- 160
- 161
- 162
- 163
- 164
- 165
- 166
注意形参里的引用与指针的合理使用!
