HDU 5418 Victor and World

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Victor andWorld***problem Descriptionafter trying hard forMany years, Victor has finally received a pilot license. To has a celebration, he intends toBuy himself an airplane andFlyaround  theWorld. There is n countries on  theEarth, which is numbered from 1  toN. They is connected byM undirected flights, detailedly theI-th Flight connects theUi-th and  theVi-th Country, and itWould cost Victor ' s airplane wi L fuelifVictor Fliesthrough it. andit  isPossible forHim toFly to everyCountry from  the  FirstCountry. Victor now is  at  theCountrywhose  Number  is 1, he wants toKnow theMinimal amount ofFuel forHim toVisiteveryCountry atLeast once andFinallyreturn  to  the  FirstCountry. Inputthe FirstLine of  theInputcontainsAnintegerT, denoting the  Number  ofTest cases. IncheveryTest case, there is integers n andMinch  the  FirstLine, denoting the  Number  of  theCountries and  the  Number  of  theFlights. Then there is m lines, each linecontainsThree integers UI, vi andWI, describing a flight.1≤t≤.1≤n≤.1≤m≤100000.1≤wi≤.1≤ui,vi≤n. Outputyour program should print T lines: theI-th ofThese shouldcontainA singleinteger, denoting theMinimal amount ofFuel forVictor toFinish theTravel. Sample Input13 21 2 21 3 3Sample OutputTen

After years of hard work, Victor finally got a flying license. To celebrate the event, he decided to buy himself a plane and travel around the world. He would fly a plane along the prescribed route. On Earth, a total of nn countries, numbering from 11 to NN, each country through the MM two-way route connection, section II route connection U_iu
? i
?? Countries and Section V_iv
? i
?? Countries, this route requires consumption of W_IW
? i
?? Oil, and from 11th countries can directly or indirectly reach 22 to any country in NN.

Victor begins in country 11th and wants to know the minimum amount of total oil consumed by countries at least once and finally returning to country 11th from country 11th.

Problem Solving Ideas:
Floyd algorithm and state DP

On the code:

/*2015-8-29 Noon Author:itak Today I want to go beyond yesterday's I, tomorrow I will surpass today's me, to create better code for the goal, constantly surpass themselves. */#include <cstdio>#include <iostream>using namespace STD;#define INF 1e9+7Const intSizet = (1<< -);intd[ -][ -];intdp[sizet][ -];intN,m;intMain () {intTscanf("%d", &t); while(t--) {scanf("%d%d", &n,&m); for(intI=0; i<n; ++i) for(intj=0; j<n; ++J) {if(i = = j) D[i][j] =0;ElseD[I][J] = INF; }intU,v,w; for(intI=0; i<m; ++i) {scanf("%d%d%d", &u,&v,&w);if(d[u-1][v-1] > W) {d[u-1][v-1] = W; d[v-1][u-1] = W; }        } for(intk=0; k<n; k++) for(intI=0; i<n; i++) for(intj=0; j<n; J + +) D[i][j] = min (d[i][j],d[i][k]+d[k][j]); for(intI=0; i<sizet; ++i) for(intj=0; j<n;        ++J) Dp[i][j] = INF; dp[1][0] =0; for(intk=1; k<sizet; ++K) for(intI=0; i<n; ++i) for(intj=0; j<n; ++J) dp[k| (1<<i) | (1&LT;&LT;J)][i] = min (dp[k| (1<<i) | (1&LT;&LT;J)][i],d[j][i]+dp[k| (1&LT;&LT;J)][j]);intMinx=inf;inttt = (1<<n)-1; for(intI=0; i<n; i++)if(Minx > dp[tt][i]+d[i][0]) Minx = dp[tt][i]+d[i][0];printf("%d\n", Minx); }return 0;}

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HDU 5418 Victor and World