Barricade(传送门)
题意
N个点M 条路径,每条路径长度为1,敌人从M节点点要进攻1节点,敌人总是选择最优路径即最短路径来进攻我方,为了阻止敌人,我们要把一些路封死,每条路径封死需要一些花费,求最小花费。
解题思路
很明显最小割,删除一些边让起点和终点不连通,接着需要解决的就是最短路,直接用ISAP算法中的层次图 dep 数组,如果它的大小大于最开始的值了,证明当前处理的就不是最短路了,我们可以直接退出或者是不将接下来求得的结果放入最终的花费当中即可
代码
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define FIN freopen("input.txt", "r", stdin)
using namespace std;
typedef long long LL;
const int MAXN = 1e4 + 5;
const int INF = 0x3f3f3f3f;
struct Edge {
int u, v, flow, cap, nxt, cost;
} E[MAXN << 1];
int Head[MAXN], cur[MAXN], tot;
int pre[MAXN], dep[MAXN], gap[MAXN];
bool vis[MAXN];
queue<int>Q;
void edge_init() {
tot = 0;
memset(Head, -1, sizeof(Head));
}
void add_edge(int u, int v, int cap) {
E[tot].u = u;
E[tot].v = v;
E[tot].cap = cap;
E[tot].flow = 0;
E[tot].nxt = Head[u];
Head[u] = tot ++;
E[tot].u = v;
E[tot].v = u;
E[tot].cap = 0;
E[tot].flow = 0;
E[tot].nxt = Head[v];
Head[v] = tot ++;
}
void BFS(int start, int end) {
memset(dep, -1, sizeof(dep));
memset(gap, 0, sizeof(gap));
gap[0] = 1;
dep[end] = 0;
while(!Q.empty()) Q.pop();
Q.push(end);
while(!Q.empty()) {
int u = Q.front();
Q.pop();
for(int i = Head[u]; ~i; i = E[i].nxt) {
int v = E[i].v;
if(dep[v] != -1) continue;
dep[v] = dep[u] + 1;
gap[dep[v]] ++;
Q.push(v);
}
}
}
int sap(int start, int end, int N) {
BFS(start, end);
memcpy(cur, Head, sizeof(Head));
memset(pre, -1, sizeof(pre));
int top = 0;
int u = start;
int ans = 0;
int Mindep = dep[start];
while(dep[start] < N) {
if(u == end) {
int Min = INF;
int inser;
for(int i = pre[u]; ~i; i = pre[E[i].u]) {
if(Min > E[i].cap - E[i].flow) {
Min = E[i].cap - E[i].flow;
inser = i;
}
}
for(int i = pre[u]; ~i; i = pre[E[i].u]) {
E[i].flow += Min;
E[i ^ 1].flow -= Min;
}
if(Mindep == dep[start])//不加入最终的花费中
ans += Min;
u = E[inser].u;
continue;
}
bool flag = false;
int v ;
for(int i = 0; i < 100; i ++)
for(int i = cur[u]; ~i; i = E[i].nxt) {
v = E[i].v;
if(E[i].cap - E[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; i = E[i].nxt) {
if(E[i].cap - E[i].flow && dep[E[i].v] < Min) {
Min = dep[E[i].v];
cur[u] = i;
}
}
gap[dep[u]] --;
if(!gap[dep[u]]) return ans;
dep[u] = Min + 1;
gap[dep[u]] ++;
if(u != start) u = E[pre[u]].u;
}
return ans;
}
int _, N, M;
int main() {
//FIN;
int a, b, c;
scanf("%d", &_);
while(_ --) {
edge_init();
scanf("%d%d", &N, &M);
for(int i = 0; i < M; i ++) {
scanf("%d%d%d", &a, &b, &c);
add_edge(a, b, c);
add_edge(b, a, c);
}
printf("%d\n", sap(1, N, N));
}
return 0;
}