Caocao‘s Bridges

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Caocao was defeated by Zhuge Liang and Zhou Yu in the battle of Chibi. But he wouldn‘t give up. Caocao‘s army still was not good at water battles, so he came up with another idea. He built many islands in the Changjiang river, and
based on those islands, Caocao‘s army could easily attack Zhou Yu‘s troop. Caocao also built bridges connecting islands. If all islands were connected by bridges, Caocao‘s army could be deployed very conveniently among those islands. Zhou Yu couldn‘t stand
with that, so he wanted to destroy some Caocao‘s bridges so one or more islands would be seperated from other islands. But Zhou Yu had only one bomb which was left by Zhuge Liang, so he could only destroy one bridge. Zhou Yu must send someone carrying the
bomb to destroy the bridge. There might be guards on bridges. The soldier number of the bombing team couldn‘t be less than the guard number of a bridge, or the mission would fail. Please figure out as least how many soldiers Zhou Yu have to sent to complete
the island seperating mission.

Input

There are no more than 12 test cases.

In each test case:

The first line contains two integers, N and M, meaning that there are N islands and M bridges. All the islands are numbered from 1 to N. ( 2 <= N <= 1000, 0 < M <= N² )

Next M lines describes M bridges. Each line contains three integers U,V and W, meaning that there is a bridge connecting island U and island V, and there are W guards on that bridge. ( U ≠ V and 0 <= W <= 10,000 )

The input ends with N = 0 and M = 0.

Output

For each test case, print the minimum soldier number Zhou Yu had to send to complete the mission. If Zhou Yu couldn‘t succeed any way, print -1 instead.

Sample Input

3 3
1 2 7
2 3 4
3 1 4
3 2
1 2 7
2 3 4
0 0

Sample Output

-1
4

题意：有n座岛和m条桥，每条桥上有w个兵守着，现在要派不少于守桥的士兵数的人去炸桥，只能炸一条桥，使得这n座岛不连通，求最少要派多少人去。

分析：只需要用Tarjan算法求出图中权值最小的那条桥就行了。

注意：

一：如果图不连通，不用派人去炸桥，直接输出0

二：可能会有重边

三：如果桥上没有士兵守着，那至少要派一个人去炸桥。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define maxn 1010
#define maxm 1000100
#define INF 0x3f3f3f3f
using namespace std;
int n, m;
int low[maxn];
int dfn[maxn];
int head[maxn], cnt;
int dfs_clock;
int mark;//判断图是否连通

struct node{
    int u, v, w, cnt, next;
};

node edge[maxm];

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int w){
    edge[cnt] = {u, v, w, 0, head[u]};
    head[u] = cnt++;
    edge[cnt] = {v, u, w, 0, head[v]};
    head[v] = cnt++;
}

void input(){
    while(m--){
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        add(u, v, w);
    }
}

void tarjin(int u, int fa){
    low[u] = dfn[u] = ++dfs_clock;
    int flag = 1;
    for(int i = head[u]; i != -1; i = edge[i].next){
        int v = edge[i].v;
        if(flag && v == fa){//去重边，重边不可能是桥
            flag = 0;
            continue;
        }
        if(!dfn[v]){
            tarjin(v, u);
            low[u] = min(low[u], low[v]);
            if(dfn[u] < low[v])//是桥
                edge[i].cnt = edge[i ^ 1].cnt = 1;
        }
        else
            low[u] = min(low[u], dfn[v]);
    }
}

void find(){
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    dfs_clock = 0;
    tarjin(1, 1);
    mark = 1;
    for(int i = 1; i <= n; ++i){
        if(!dfn[i]){
            mark = 0;
            return ;
        }
    }
}

void solve(){
    if(!mark)
        printf("0\n");
    else{
        int ans = INF;
        for(int i = 0; i < cnt; ++i){
            if(edge[i].cnt)
                ans = min(ans, edge[i].w);
        }
        if(ans == INF)
            ans = -1;
        if(ans == 0)
            ans = 1;
        printf("%d\n", ans);
    }
}

int main (){
    while(scanf("%d%d", &n, &m), n || m){
        init();
        input();
        find();
        solve();
    }
    return 0;
}

另一种去重边的方式，对这一题来说处理时间比较慢

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#define maxn 1010
#define maxm 1000100
#define INF 0x3f3f3f3f
using namespace std;
int n, m;
int low[maxn];
int dfn[maxn];
int head[maxn], cnt;
int dfs_clock;
int mark;//判断图是否连通

struct node{
    int u, v, w, cnt, next, again;//again标记是不是有重边
};

node edge[maxm];

void init(){
    cnt = 0;
    memset(head, -1, sizeof(head));
}

void add(int u, int v, int w){
    int i;
    //标记重边
    for(i = head[u]; i != -1; i = edge[i].next){
        if(edge[i].v == v){
            break;
        }
    }
    if(i != -1)
        edge[i].again = 1;
    else{
        edge[cnt] = {u, v, w, 0, head[u], 0};
        head[u] = cnt++;
    }
}

void input(){
    while(m--){
        int u, v, w;
        scanf("%d%d%d", &u, &v, &w);
        add(u, v, w);
        add(v, u, w);
    }
}

void tarjin(int u, int fa){
    low[u] = dfn[u] = ++dfs_clock;
    for(int i = head[u]; i != -1; i = edge[i].next){
        int v = edge[i].v;
        if(v == fa) continue;
        if(!dfn[v]){
            tarjin(v, u);
            low[u] = min(low[u], low[v]);
            if(dfn[u] < low[v] && edge[i].again == 0)//是桥
                edge[i].cnt = edge[i ^ 1].cnt = 1;
        }
        else
            low[u] = min(low[u], dfn[v]);
    }
}

void find(){
    memset(dfn, 0, sizeof(dfn));
    memset(low, 0, sizeof(low));
    dfs_clock = 0;
    tarjin(1, -1);
    mark = 1;
    for(int i = 1; i <= n; ++i){
        if(!dfn[i]){
            mark = 0;
            return ;
        }
    }
}

void solve(){
    if(!mark)
        printf("0\n");
    else{
        int ans = INF;
        for(int i = 0; i < cnt; ++i){
            if(edge[i].cnt)
                ans = min(ans, edge[i].w);
        }
        if(ans == INF)
            ans = -1;
        if(ans == 0)
            ans = 1;
        printf("%d\n", ans);
    }
}

int main (){
    while(scanf("%d%d", &n, &m), n || m){
        init();
        input();
        find();
        solve();
    }
    return 0;
}

版权声明：本文为博主原创文章，未经博主允许不得转载。

HDU 4738 --Caocao's Bridges 【无向图边双联通 && 求权值最小的桥 && 模板】