【codeforces 716D】Complete The Graph

【题目链接】:http://codeforces.com/problemset/problem/716/D

【题意】



给你一张图;

这张图上有一些边的权值未知;

让你确定这些权值(改成一个正整数)

使得s到t的最短路恰好为L

【题解】



首先;

算出两个值

temp1->所有的未知边的权值都为1->算出s到t的最短路;

temp2->所有的未知边的权值都为INF->算出s到t的最短路;

则必须要有

temp1<=L<=temp2

否则无解;

明白这个之后;

为每一个未知的边都标号;

标号为1..totl;

然后;

二分有多少条未知边的权值边为1;

->mid

找到最小的,使得在mid条未知边的权值为1的时候;

s到t的最短路小于L;

则第mid条边必然在s->t的最短路上;

则把那第mid条边再加上s->t的最短路与L的差值

(前Mid-1条边权值还是1);

(因为1的边越多,s到t的最短路是单调不上升的,所以这么做是可行的)

(又因为是>L和< L的边界,所以那个mid一定是在最短路上的,且没有它最短路就会大于L)



【Number Of WA】



1



【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ms(x,y) memset(x,y,sizeof x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const double pi = acos(-1.0);
const int N = 1100;
const int M = 20000+100;
const int INF = 1e9+1;

int fir[N],nex[M],en[M],w[M],lable[M];
int totm,totl,n,m,L,s,t;
LL dis[N];
bool exsit[N];
queue <int> dl;

void add(int x,int y,int z,int flag)
{
    nex[totm] = fir[x];
    fir[x] = totm;
    en[totm] = y;
    w[totm] = z;
    if (flag) lable[totm] = totl;
    totm++;
}

LL spfa(int pre)
{
    rep1(i,1,n) dis[i] = -1;
    exsit[s] = true;
    dl.push(s);
    dis[s] = 0;
    while (!dl.empty())
    {
        int x = dl.front();
        dl.pop();
        exsit[x] = false;
        for (int i = fir[x];i>=0;i = nex[i])
        {
            int y = en[i],cost = w[i];
            if (lable[i] && lable[i]<=pre) cost = 1;
            if (lable[i] && lable[i]>pre) cost = INF;
            if (dis[y]==-1 || dis[y]>dis[x]+cost)
            {
                dis[y] = dis[x]+cost;
                if (!exsit[y])
                {
                    exsit[y] = true;
                    dl.push(y);
                }
            }
        }
    }
    return dis[t];
}

void out_graph(int key,int sp )
{
    rep1(i,1,n)
    {
        for (int j = fir[i];j >= 0;j = nex[j])
        {
            if (j&1) continue;
            cout <<i-1<<' '<<en[j]-1<<' ';
            int cost = w[j];
            if (lable[j])
            {
                if (lable[j]<key) cost = 1;
                if (lable[j]==key) cost = sp;
                if (lable[j]>key) cost = INF;
            }
            cout << cost << endl;
        }
    }
}

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    ios::sync_with_stdio(false),cin.tie(0);//scanf,puts,printf not use
    cin >> n >> m >> L >> s >> t;s++,t++;
    rep1(i,1,n) fir[i] = -1;
    rep1(i,1,m)
    {
        int x,y,z;
        cin >> x >> y >> z;
        x++,y++;
        if (z==0) totl++;
        add(x,y,z,z==0);
        add(y,x,z,z==0);
    }
    LL temp1 = spfa(totl),temp2 = spfa(0);
    if (temp1 <= L && L <=  temp2)
    {
        cout << "YES" << endl;
        int l = 0,r = totl,ans;
        LL tl,ansl;
        while (l <= r)
        {
            int mid = (l+r)>>1;
            tl = spfa(mid);
            if (tl<=L)
                ans = mid,r = mid-1,ansl = tl;
            else
                l = mid+1;
        }
        out_graph(ans,L-ansl+1);
    }
    else
        cout << "NO" << endl;
    return 0;
}